HPSC Assistant Professor 2018 Physics (Official Paper) (Previous Year Paper)

Concept:

• Refractive Index (μ):  It is the ratio of the speed of light in vacuum (or air, approximately) to the speed of light in the medium. A higher refractive index indicates that light travels slower in that medium.

         $\mu =v/c$

Optical Path Length (OPL): 

• The optical path length is the product of the actual physical path length a light wave travels and the refractive index of the medium through which it propagates. Hence, for a medium with refractive index μ and thickness t, the optical path length is given by:

         $\text{Optical Path} = \mu \cdot t$

Explanation:

When a thin film of thickness t and refractive index μ is placed in the path of a light wave emerging from a source such as S1, the optical path length is altered due to the change in the medium through which the light travels.

• Without the Medium: The refractive index of air is approximately 1).

  $\text{Optical Path}_{\text{without medium}} = 1.t = t$

• With the Medium: The refractive index of medium is μ. So, the new optical path length will be,

  $\text{Optical Path}_{\text{with medium}} = \mu \cdot t$

There is an  increase in the optical path length (ΔOptical Path) due to the presence of the thin film which is equal to the difference between the optical path lengths with and without the medium:

\( \Delta \text{Optical Path} = \text{Optical Path}{\text{with medium}} - \text{Optical Path}{\text{without medium}} \)

$\Delta \text{Optical Path} = (\mu \cdot t) - t$

$\Delta \text{Optical Path} = t(\mu - 1)$

The correct answer is (μ - 1) t.

Concept:

Unpolarised Light: The electric field vector for an unpolarised light oscillate in random direction.

Polarised Light: The electric field vector has specific direction in which it propagates. It can be linearly polarised, circularly polarised etc.

Polarizer: When we passes an unpolarised light through a polariser the it gets polarised that means that it only allows specific directions of electric field to pass through it and other components are blocked. This means that the intensity is also reduced as not all the components are allowed to pass through it.

Malus's Law: Since the polariser allows few components to pass through it that means the intensity is reduced as not all the components are allowed to pass through it.​

The Malus's Law related the intensity of light incident on the polarizer before and after it passes through it.

  $I =I_0 \cos^2 θ$

Where,$I_0$ is the initial intensity before it passes through the polariser and $I$ is the final intensity of light after it passes through the polariser. θ is the angle between the axis of polariser and the light polarisation direction.

Explanation: 

Since the polariser is rotating so the value of θ can be from $0$ to $2\pi$. We can now find out the average intensity,

$I_\text{avg}= \frac{1}{2\pi}\int_0^{2\pi} I_0 \cos^2\theta d\theta\ \ \ \ \ \ \ = \frac{1}{2\pi}\int_0^{2\pi} I_0 \frac{(1-\cos2\theta)}{2} d\theta \ \ \ \ \ \ \ =\frac{I_0}{2}$

Time period of revolution is  $(\text{T})=\frac{2\pi}{ω}$.

$\text{Intensity}= \frac {\text{Energy}} { \text{Area × Time Period}}\ I_0 = \frac{10^{-3}}{3\times 10^{-4}\times 1}=\frac{10}{3} \frac{ \text{watt}}{ \text{m}^2}$

Given value of ω is  31.4 rad/s.

$\text{T}= \frac {2 × 3.14}{31.4} \ \ \ \ =0.2\text{sec}$

Energy of light passing through the polariser is = Average intensity × Area × Time Period of revolution.

$\text{E}= \frac{10}{3\times 2} \times 3\times 10^{-4} \times 0.2 \ \ \ \ = 10^{-4}\text{J}$

The correct answer is 10-4 J.

Concept:

There are four Maxwell's equations as below:

1. Gauss's Law of Electricity:  $\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$ 2.  Gauss's Law of Magnetism:  $\nabla \cdot \mathbf{B} = 0$ 3. Faraday's Law of Induction: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$ 4. Ampere's Law (along with Maxwell's corrections): $\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$

Explanation:

From here if we use electric field and electric potential relation in 1^st equation  which is $\mathbf{E}=-\nabla \cdot \mathbf{V}$. We get the following 

$\nabla^2 \cdot \mathbf{V} = -\frac{\rho}{\epsilon_0}$

This equation is the Poisson's equation. In cgs it can be written as below:

$\nabla^2 \cdot \mathbf{V} = -4\pi\rho$

The correct answer is Poisson's equation.

Concept:

Electromagnetic waves have both electric and magnetic field components which are perpendicular to each other. Hence, the total energy density for electromagnetic waves will have contribution from both electric and magnetic fields.

The total energy density ( u ) of the electromagnetic wave is given by:

 $u =u_e+u_m= \frac{1}{2} ( ϵ_0 E^2 + \frac{1}{μ_0} B^2 )$ 

where: ϵ 0 is the permittivity of free space and its value is  $8.854 \times 10^{-12} \text{F/m}$}.

μ0 is the permeability of free space and its value is $4\pi \times 10^{-7} \text{H/m}$.

E is the electric field strength.

B is the magnetic field strength.

For a plane electromagnetic wave, the electric and magnetic fields are related by $B = \frac{E}{c}$ , where c is the speed of light in a vacuum ( $c \approx 3 \times 10^8 \text{m/s}$ ). Given this relationship, the energy density can be rewritten as: [$u = ϵ_0 E^2$  ]

Hence, Option 4 is the correct answer.

Concept:

When a magnetic dipole is placed in a magnetic field, it experiences a torque that will try to align the dipole with the field. This torque is due to the interaction between the magnetic moment of the dipole and the magnetic field in which it is placed.

For our case, the magnetic field is non-uniform, the dipole will experience a net translational force along . In this context, we will derive the expression for the torque acting on a magnetic dipole in a non-uniform magnetic field.

The torque ( $\vec{\tau}$ ) on a magnetic dipole ($\vec{m}$) placed in a magnetic field ( $\vec{B}$ ) is given by:

$\vec{\tau} = \vec{m} \times \vec{B}$

Here, $\vec{m}$  is the magnetic dipole moment vector, and $\vec{B}$  is the magnetic field vector.

As we know that a dipole experience a torque when placed in an electric field. Similarly, the torque for magnetic field will try to align the dipole moment with the magnetic field. 

If the magnetic field is non-uniform, the magnetic dipole can also experience a force. This force ( $\vec{F}$ ) is given by taking the gradient of the potential energy ( $U = -\vec{m} \cdot \vec{B}$ ) of the dipole in the magnetic field:

$\vec{F} = \vec{\nabla} (\vec{m} \cdot \vec{B})$

Given that the position vector of the dipole is $\vec{r}$ , the torque about the origin can be expressed as the cross product of the position vector ( $\vec{r}$ ) and the force ( $\vec{F}$ ):

$\vec{\tau}_{\text{origin}} = \vec{r} \times \vec{F}$

Explanation:

The force ( $\vec{F}$ ) in terms of the position vector ( $\vec{r}$ ) is :

$\vec{F} = \vec{\nabla} (\vec{m} \cdot \vec{B})$

Therefore, the torque about the origin is:

$\vec{\tau}_{\text{origin}} = \vec{r} \times \left[ (\vec{m} \cdot \vec{\nabla}) \vec{B} \right]$

Summarizing, the torque acting on a magnetic dipole in a non-uniform magnetic field, considering the position vector ( $\vec{r}$ ) from the origin after solving, is given by the expression:

$\vec{\tau}_{\text{origin}} =(\vec{m}\times\vec{B})+ \vec{r} \times \vec{\nabla} (\vec{m} \cdot \vec{B} )$

This torque tends to rotate the dipole in the magnetic field and can also depend on the position of the dipole in the non-uniform field.

Hence, the correct answer is option 2.

Concept:

To satisfy the Coulomb Gauge the following equation must be satisfied:

$\vec{\nabla}\cdot \vec{A}=0$

For Lorentz Gauge, the condition is:

$\vec{\nabla}\cdot \vec{A}+\epsilon_0\mu_0 \frac{\partial \phi}{\partial t}=0$

Explanation:

Given, the vector potential is

$\vec{A}(\vec{x}, t) = (2x - ω t)\hat{\mathbf{i}} + (y - 2z)\hat{\mathbf{j}} + (z - 2e^{iω t})\hat{\mathbf{k}}$

We can find out whether it satisfies the Coulomb gauge or not,

$\vec{\nabla}\cdot \vec{A} = [\frac{\partial }{\partial x}\hat{\mathbf{i}}+\frac{\partial }{\partial y}\hat{\mathbf{j}}+\frac{\partial }{\partial z}\hat{\mathbf{k}} ] \cdot (2x - ω t)\hat{\mathbf{i}} + (y - 2z)\hat{\mathbf{j}} + (z - 2e^{iω t})\hat{\mathbf{k}} \ \ \ \ \ \ \ \ \ \ = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}\\ \ \ \ \ \ \ \ \ = \frac{\partial (2x - ω t)}{\partial x} + \frac{\partial (y - 2z)}{\partial y} + \frac{\partial (z - 2e^{iω t})}{\partial z} =2+1+1=4\neq 0$

Given, the scalar potential is 

\(ϕ (x̅, t) = 3xyz - 4t \ \frac{\partial \phi (x̅, t)}{\partial t}=\frac{\partial (3xyz - 4t) }{\partial t}=-4 \)

Hence,

$\vec{\nabla}\cdot \vec{A}+\epsilon_0\mu_0 \frac{\partial \phi}{\partial t}\neq0$

So, it will not satisfy both Coulomb and Lorentz Gauge.

For, ω =0 also both the equations are not satisfied as they have constant values after Coulomb and Lorentz Gauge.

So, these potentials do not describe any electromagnetic field.

The correct option is option 4.

Concept:

Surface Current Density: Surface current density (K) is defined as the current per unit length flowing along a certain direction on a surface. In our case, infinite conducting sheet, this is represented as:

$[ \vec{K} = K_y \hat{\mathbf{y}} ]$

where, Ky is the magnitude of the surface current density along the (y)-axis, and $(\hat{\mathbf{y}})$ is the unit vector in the (y) direction.

This presence of a surface current density ( $\vec{K}$) on the conductive sheet induces both electric and magnetic fields around it. We can analyze these fields using the boundary conditions that result from Maxwell's equations.

Explanation:

Since the current is infinite in extent and flows uniformly, we use Ampere's Law with the surface integral form:

$\oint \vec{B} \cdot d\vec{l} = μ_0 I_{\text{enc}}$

 where, μ₀ is the permeability of free space.

$I_{\text{enc}} = K_y l$

$\oint \vec{B} \cdot d\vec{l} = \int_{AB} \vec{B} \cdot d\vec{l}+ \int_{BC} \vec{B} \cdot d\vec{l}+ \int_{CD} \vec{B} \cdot d\vec{l}+ \int_{DA} \vec{B} \cdot d\vec{l}$

Since, BC and CD re perpendicular to the magnetic field and direction of current, So they will have zero contribution.

While AB  and DA are in opposite directions but both are

$\oint \vec{B} \cdot d\vec{l} = 2Bl =\mu_0 K l \ B=\frac{\mu_0 K}{2}$

By using right hand thumb rule we can find out the direction of magnetic field.

The magnetic field direction is in +x.

So, the correct answer is option 3.

Concept:

The vector potential for a particular current distribution is given by the following relation

$\vec{A}=\frac{\mu_0}{4\pi}\int \frac{Jdτ}{r}$

Where, J is the current density.

dτ is the volume element and r is the distance at which we are interested to find out the magnetic field.

Explanation:

Here, the current distribution is not defined. So we can not find out the value of Vector potential.

The correct answer is option 4.

Concept:

Michelson Interferometer: It is used for measuring very small distances, wavelengths of light, and the refractive indices of various materials. It is based on the principle of interference, where waves, such as light waves, can constructively or destructively interfere with each other. Hence, depending on the path difference there are bright and dark fringes.

This path difference changes when we change the position of mirror. This will also affect the fringes formation. The following relation relates how the changes in the position of mirror(Δ x) is related to the fringes.

$2 Δ x= nλ$

Where, n is the number of fringes and λ is the wavelength of monochromatic light.

Explanation:

Given,

Δ x= 25.3μm; n= 92

\(\lambda=\frac{2\Delta x}{n}=\frac{2 \times 25.3 \mu m}{92}

\ \lambda=550 \text{nm}\)

The correct answer is option 2.

Concept:

By Gauss's Law, we have 

$\oint \overrightarrow E \cdot \overrightarrow S =\frac{Q_{enc}}{\epsilon_0}$

Where, $\overrightarrow E$ is the electric field and $\overrightarrow S$  is the surface of integration over which we need to find out the electric field.

$E \oint S =\frac{q}{\epsilon_0} \ E=\frac{q}{4\pi r^2 \epsilon_0}$

where, r is the distance inside the sphere from its centre.

Explanation:

The charge density ρ is given by:

$\rho = \frac{Q}{\frac{4}{3} \pi R^3}\$

For a distance d from the center, the enclosed charge $Q_{\text{enc}}\$ is:

$Q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi d^3 = \frac{Q}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi d^3 = Q \left( \frac{d^3}{R^3} \right)\$

By Gauss's Law, the electric flux is:

$\oint \vec{E} \cdot d\vec{A} = E \cdot 4 \pi d^2 \$

Thus:

$E \cdot 4 \pi d^2 = \frac{Q \left( \frac{d^3}{R^3} \right)}{\epsilon_0} \$

Solving for the electric field E, we get:

$E = \frac{Q d}{4 \pi \epsilon_0 R^3}\$

Let $k = \frac{Q}{4 \pi R^3}\$, then:

$E = \frac{kd }{\epsilon_0}\$

The direction of the electric field is radial, denoted by $\hat{r}\$.

Concept:

 The mass of an electron in relativity changes with the speed. The formula is given as:

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Solution:

Since the mass becomes two times the intial mass we can use this formula here,

\(m=2m_0 \ \ 2=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \

\ 1-\frac{v^2}{c^2}=\frac{1}{4}

\ \ \frac{v^2}{c^2}=\frac{3}{4}

\ \ v=\frac{\sqrt 3}{2}c

\)

So, the correct answer is option1.

Explanation:

The electric and magnetic field are frame dependent. For example, if we have a frame S' which is moving with velocity v with respect to frame S which is at rest. Then, the electric and magnetic fields will transform as:

E_x' = E_x

E_y' = γ (E_y - v B_z)

E_z' = γ (E_z + v B_y)

B_x' = B_x

B_y' = γ (B_y + v / c² E_z)

B_z' = γ (B_z - v / c² E_y)

Where, prime coordinates are in S' frame.

The electromagnetic field in relativistic physics is conveniently described using the electromagnetic field tensor ( F_μν) which is a 4x4 antisymmetric matrix that combines the electric field and the magnetic field. In Cartesian coordinates (with the speed of light (c) taken into account appropriately), it can be written as:

F_μν = E - c² B

This quantity is invariant from any frame.

The correct answer is option 2.

Concept:

The relation between ω and velocity(v) is,

$\text{v}=\frac{d\omega}{dk}$

Explanation: 

Given,  

$\omega^2=ka \ \text{v}_g=\frac{d\omega}{dk}\ 2\omega d\omega=a dk\ \frac{d\omega}{dk}=\frac{a}{2\omega} \ \text{v}_g=\frac{a}{2\omega}$

The correct answer is option 1.

Concept:

By Gauss's law, we have the divergence of the electric field related to the charge density of the distribution(ρ ).

$\nabla \cdot \mathbf{E}=\frac{ρ}{\epsilon_0}$

Explanation:

The divergence in Spherical polar coordinates is given as:

$\nabla \cdot \mathbf{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta E_\theta) + \frac{1}{r \sin \theta} \frac{\partial E_\phi}{\partial \phi}$

Given electric field:

$\mathbf{E}(\vec{x})=\frac{qe^{-ar}}{r^3} \hat{r} \$

Which has  only the radial component. Hence, only the radial term will contribute for the divergence.

$\nabla \cdot \mathbf{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2\cdot \frac{qe^{-ar}}{r^3} ) \ \nabla \cdot \mathbf{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (\frac{qe^{-ar}}{r} ) \ \nabla \cdot \mathbf{E} = \frac{1}{r^2}[-\frac{qae^{-ar}}{r}-\frac{qe^{-ar}}{r^2}]\ \frac{\rho}{\epsilon_0 }=-\frac{q e^{-ar}}{r^4}(ar+1) \ \rho=-\frac{q\epsilon_0 e^{-ar}}{r^4}(ar+1)$

Hence, option 2 is the correct answer.

Concept:

In photoelectric effect, when light is incident on some metal surface then there is an ejection of electrons takes place if the light has energy more than the work function of that metal.

The work function is the minimum energy required to eject electrons from a metal.

The equation which relates the energy of the ejected electrons to the work function and the incident energy is:

$\text{E}_k=\frac{hc}{λ}-\phi_0$

Where, Ek is the maximum kinetic energy of the ejected electrons.

λ is the wavelength of incident light.

$\phi_0$  is the work function of the metal.

Explanation:

Given, the threshold wavelength(λ0 ) of tungsten is 2300A°.

The wavelength of the incident light(λ ) is 1800A°.

Work function of tungsten is,

$\phi_0 =\frac{hc}{\lambda_0}= \frac{6.626 \times 10^{-34}Js\times 3\times 10^8 ms^{-1}}{2300\times 10^{-10}m}\ \phi_0\approx 5.39\text{eV}$

The incident energy=$\frac{hc}{\lambda} =\frac{6.626 \times 10^{-34}Js\times 3\times 10^8 ms^{-1}}{1800\times 10^{-10}m}\ \phi_0\approx 6.88 \text{eV}$

Hence,

$\text{E}_k=(6.88-5.39)\text{eV}\approx 1.5\text{eV}$

The correct answer is option 2.

Concept:

The change in mass for relativistic case is given by the formula as:

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Solution:

Since the mass becomes two times the intial mass we can use this formula here,

\(v=0.8c \ \ m=\frac{m_0}{\sqrt{1-\frac{0.64c^2}{c^2}}} \m=\frac{m_0}{\sqrt{1-0.64}} \m=\frac{m_0}{\sqrt{0.36}} \ m=\frac{5}{3}m_0 \)

So, the correct answer is option 3.

In photoelectric effect, when light is incident on some metal surface then there is an ejection of electrons takes place if the light has energy more than the work function of that metal.

The work function is the minimum energy required to eject electrons from a metal.

The equation which relates the energy of the ejected electrons to the work function and the incident energy is:

$\text{E}_k=\frac{hc}{λ}-\phi_0$

Where, Ek is the maximum kinetic energy of the ejected electrons.

λ is the wavelength of incident light.

$\phi_0$  is the work function of the metal.

The intensity is the energy per unit area per unit time. The more the intensity more the number of electrons will be emitted.

Solution:

• The maximum kinetic energy of the photoelectrons is independent of the light intensity. 

 E ∝ 1/λ

• The number of electrons emitted is proportional to the intensity of the light. Also, illumination is nothing but the emission of light from a surface when light is incident on it. Which is dependent on the intensity.

N ∝ L 

So, the correct answer is option 2.

Concept:

Bragg's law gives the condition under which a diffracted beam can be observed for a set of parallel lattice planes with indices h,k,l of spacing d and at an angle of incidence θ, when a beam of parallel X-rays incident on it.

$\text{n}λ = 2\text{d} \sin θ$

The value of Sine function can be between -1 to +1.

So,  $2\text{d}\sinθ < 2\text{d} \ \text{Hence,} \ \ \text{n}λ <2\text{d} \ \text{for n=1}; \ λ <2\text{d} \ \text{for n=2}; \ λ <\text{d}$

This is the condition for Bragg's Law to follow.

λ =d, can also be true for n=2 case when θ =90^o.

The only option which does not follow these conditions is option 1.

So, the correct answer is option 1.

Concept:

Momentum of a particle: It is the product of mass(m) and velocity(v) of a particle and for waves it is the ratio of Plancks constant and the wavelength(λ ) associated with it.

$p=m\cdot v =\frac{h}{λ}$

Explanation:

Given,

\(m=200\text{g}=0.2\text{kg}\

v=5 \text{m/h}= \frac{5}{3600}\text{m/sec}\ \lambda=\frac{h}{mv}=\frac{ 6.626 × 10-34 \text{Js}}{0.2\text{kg}\times \frac{5}{3600}\text{m/sec} } \

\lambda \approx 10^{-30}\text{m}\)

The correct answer is option 3.

Concept:

Radiation can behave like both as a wave and as a particle from the dual nature of light. For example in double slit experiment we observe a wave like patterns since the wavelength is comparable to the size of the slit width. 

Compton Scattering: It explains the interaction of light with the matter. It is based on the principle that when a photon collides with a free or loosely bound electron, the photon is scattered in a different direction, and its wavelength increases (which means its energy decreases). This scattering process can be explained by using both conservation of energy and conservation of momentum.

This change in the wavelength (Δλ) of the photon due to Compton scattering depends on the angle (θ) at which the photon is scattered and is given by the Compton wavelength shift equation:

$Δ λ = λ' - λ = \frac{h}{m_e c} (1 - \cos θ)$

where, λ is the initial wavelength of the photon.

λ' is the wavelength of the scattered photon.

h is Planck's constant $6.626 \times 10^{-34} \text{Js}$ .

m_e is the rest mass of the electron $9.1 \times 10^{-31} \text{kg}$ .

(c) is the speed of light in vacuum $3 \times 10^8 \text{m/s}$

(θ) is the angle at which the photon is scattered.

The term $(\frac{h}{m_e c})$  is known as the Compton wavelength of the electron, approximately equal to $2.43 \times 10^{-12} \text{m}$ .

Explanation:

Here in Compton Scattering the interaction between the photon and electron is explained such that both behave like a particle. Since it has used conservation of momentum and energy as a particle case to find out the change in wavelength.

The correct answer is option 2.

Concept:

Bragg's Law gives the condition when X-rays fall on the surface of a lattice with particular planes which have some interplanar distance(d).

The condition for constructive interference is given as:

$\text{n}λ =2\text{d} \sinθ$

Where, n is the order of maxima.

d is the interplanar distance.

λ is the incident X-ray wavelength.

θ is the angle at which rays are falling.

Explanation:

The momentum is defined as the ratio of Planck's constant(h) to the wavelength(λ).

$\text{p}=\frac{\text{h}}{λ}$

Given, momentum is 

$\text{p}=\frac{\text{h}}{d}$

that means that λ is equal to d.

Using this in the Bragg's equation,

$\text{n}λ =2\text{d} \sinθ \ \text{n}\text{d} =2\text{d} \sinθ \$

for n=1,

$\sinθ =\frac{1}{2}\ \theta =\frac{\pi}{6}$

The correct answer is option 1.

Concept:

X-Ray Emission: X-ray emission is the process by which X-rays are generated, when an electron electron get enough energy to get excited to a higher energy level which creates a vacancy at that place and that vacant place is filled by another K-Shell or outer shell electron.

The correct answer is option 4.

Concept:

The force due to electric field(E) on a charge q is given as,

F=qE

And by newton's second law, Force is the rate of change of momentum or equal to the product of mass and acceleration.

$\text{F}=\frac{d\text{p}}{dt}=\text{ma}$

Momentum is equal to the ratio of Plancks constant and the wavelength.

$\text{p}=\frac{h}{\lambda}$

Explanation:

Since, q=-e

$|\text{F}|=e\text{E}=\text{ma}\ a=\frac{e\text{E}}{\text{m}}$

Motion with uniform velocity follows the relation:

v=u+at

where, v is the final velocity.

u is the initial velocity.

a is the acceleration.

t is the time.

$v=0+\frac{e\text{E}}{m}t\ v=\frac{e\text{E}}{m}t$

velocity is the rate of change of position. Here, it can be written as rate of change of wavelength,

$mv=\frac{h}{\lambda}\ \lambda=\frac{h}{m\cdot \frac{e\text{E}}{m}t}\ \lambda=\frac{h}{e\text{E}t}\ \frac{d\lambda}{dt}=-\frac{h}{e\text{E}t^2}$

The correct answer is option 1.

Concept:

The Schrodinger's equation for a particle in a box for 1D is,

$-\dfrac{\hslash^2}{2m} , \dfrac{\mathrm{d}^2 \psi(x)}{\mathrm{d} x^2}=\text{E}\psi(x)$

So, the wave function is 

$\psi(x)=\sqrt{\frac{2}{\text{L}}}\sin{(\frac{n\pi x}{\text{L}})}$

Where, L is the length of the box.

Explanation:

The energy eigenvalues are 

$\text{E}_n=\frac{n^2\pi^2\hslash^2}{2m\text{L}^2}$

Where, n is the principle quantum number such that it can take values such as 1,2,3,...

The correct answer is option 2.

Concept:

The expectation value of any operator in quantum mechanics with normalised wavefunction ψ is defined as:

\(<A>=\int ψ^*(\vec{x}) Aψ(\vec{x})\)

Where, A is the operator and ψ^* is the complex conjugate of the wavefunction.

Similarly, the expectation value for momentum operator is,

\(<p>=\int ψ^(\vec{x}) \ p \ ψ(\vec{x})=\int ψ^(x) \ (-i \hslash \frac{\partial}{\partial x} ) \ ψ(x)dx\)

Explanation:

Given, 

​​$\rm \psi (x)=Ne^{\frac{-x^2}{2a^2}+ikx}$

This is not a normalised wavefunction so we will normalize it by using,

$\int \psi^*(x)\psi(x) dx=1\ \int(Ne^{\frac{-x^2}{2a^2}-ikx})(Ne^{\frac{-x^2}{2a^2}+ikx}) dx=1 \ N^2 \int e^{\frac{-x^2}{a^2}}dx=1 \ N^2=\frac{1}{\int e^\frac{-x^2}{a^2}dx} =\sqrt{\frac{1}{\pi a^2}}$

Hence, the normalised wavefunction is 

$\rm \psi (x)=(\frac{1}{\pi a^2})^{1/4}e^{\frac{-x^2}{2a^2}+ikx}$

Hence, the expectation value will be,

\(<p>=\int ({\frac{1}{\pi a^2}})^{1/4} \cdot e^{-\frac{x^2}{2a^2}-ikx}\ (-i \hslash \frac{\partial}{\partial x} ) \ ({\frac{1}{\pi a^2}})^{1/4} \cdot e^{-\frac{x^2}{2a^2}+ikx} \ <p>=\frac{-i \hslash}{\sqrt{\pi a^2}} \int (e^{-\frac{x^2}{2a^2}-ikx})\ \cdot e^{-\frac{x^2}{2a^2}+ikx}(\frac{-x}{a^2}+ik)dx\ <p>=\frac{-i \hslash}{\sqrt{\pi a^2}} \int e^{-\frac{x^2}{a^2}}(\frac{-x}{a^2}+ik)dx\ <p>=\frac{-i \hslash}{\sqrt{\pi a^2}}[ \int e^{-\frac{x^2}{a^2}}\cdot \frac{-x}{a^2}dx+\int e^{-\frac{x^2}{a^2}}\cdot ikdx]\)

The first term in integral is zero as it is an even function.

\( <p>=0+\frac{-i \hslash}{\sqrt{\pi a^2}}(ik\sqrt{\pi a^2})\ <p>=\hslash k \)

Hence, the correct answer is option 1.

Concept:

The scattering cross section at low energies for a rigid sphere can be analysed using principles of quantum scattering theory. In the low-energy regime, where the wavelength of the incoming particle is much larger than the radius of the sphere, the s-wave (l = 0) scattering dominates the behaviour.

Explanation:

For a rigid sphere (or hard sphere) of radius (a), in the low-energy or long-wavelength limit, the phase shift for the s-wave, δ₀, approaches a value such that:

$δ_0 \approx -ka \$

where, k is the wave number of the incoming particle.

The total scattering cross-section $\sigma_{\text{total}} \$ at low energies is dominated by the s-wave contribution, given by:

$\sigma_{\text{total}} = \frac{4\pi}{k^2} \sin^2(\delta_0) \ \$

Substituting $δ_0 \approx -ka \$:

$\sin(δ_0) \approx \sin(-ka) \approx -ka \ \$

Thus, the total scattering cross-section for a rigid sphere of radius a at low energies is:

$\sigma_{\text{total}} = \frac{4\pi}{k^2} (ka)^2 = 4\pi a^2 \ \$

Hence, the correct answer is option 4.

Concept:

Depending on the spin there are two categories of elementary particles: Bosons and Fermions. 

Bosons: Bosons are those elementary particles whose spins have integer values like 0,1,2 etc. They obey Bose-Einstein Statistics for distribution.

e.g. photons, bosons, gravitons etc.

Fermions: Fermions are those elementary particles whose spins have half integer values like 1/2, 3/2 etc. This follows Fermi-Dirac Distribution. No two electrons with same spin can occupy same state.

e.g. electrons, protons etc.

Explanation:

The total spin for Neutral helium atom  is zero. It has 2 protons and two electrons which occupies same states with opposite spin. Hence, it is a boson.

Similarly, α — particles has zero total spin due to which it is a boson.

Photons also have zero spin. Hence it is also a boson.

Muons are having half integral spin. So, they are not bosons.

The correct answer is option 4.

Concept:

For a quantum harmonic oscillator, the probability of finding a particle outside its classical turning points can be calculated using quantum mechanics. In the classical case, a particle oscillates between two points - $x_{\text{min}} \$ and $x_{\text{max}} \$ - but in quantum mechanics, there's a non-zero probability to find the particle outside these turning points due to the tunnelling effect.

The wavefunction for the ground state (n=0) of a harmonic oscillator is Gaussian and falls off exponentially outside the classical turning points. The probability can be estimated by integrating the square modulus of the wavefunction outside these points.

Explanation:

In the ground state (n=0) of a quantum harmonic oscillator, the wavefunction is given by:

$\Psi_0(x) = \left( \frac{m ω}{\pi \hbar} \right)^{1/4} e^{- \frac{m ω x^2}{2 \hbar }} \ \$

Here, $\hbar \$ is the reduced Planck's constant, ω  is the angular frequency, and m is the mass of the particle.

The classical turning points are at $x = \pm \sqrt{\frac{2E}{m ω^2}} \ \$,

where ( E ) is the total energy of the oscillator. For the ground state, the energy is $E = \frac{1}{2} \hbar ω \$, so the turning points are:

$x = \pm \sqrt{\frac{2E}{m ω^2}} \ \$

The probability of finding the particle outside these classical limits for the ground state can be approximated by considering the tail regions of the Gaussian wave function. This is calculated by integrating the square of the wavefunction from the classical turning point to infinity.

$P(x > x_{\text{classical}}) + P(x < - x_{\text{classical}}) \approx 2\int_{ x_{\text{classical}} }^\infty \Psi_0(x)^2 dx \ \$

Then approximated probability often found: around 0.158.

Hence option 1 is the correct answer.

Concept:

The following are the Pauli matrices σ x, σ y ,σ z .

 σ x  = $\begin{bmatrix} 0 & 1\ 1 & 0 \end{bmatrix}$  ,  σ y  = $\begin{bmatrix} 0 &-i\ i & 0 \end{bmatrix}$   ,  σ z = $\begin{bmatrix} 1& 0\ 0 & -1 \end{bmatrix}$  

Calculation:

 σ x  σ y = $\begin{bmatrix} 0 & 1\ 1 & 0 \end{bmatrix}$  $\begin{bmatrix} 0 &-i\ i & 0 \end{bmatrix}$  =$\begin{bmatrix} i &0\ 0 & -i \end{bmatrix}$ =$i \begin{bmatrix} 1& 0\ 0 & -1 \end{bmatrix}$ =  iσ z

The correct option is 3.

Concept:

The commutator bracket in quantum mechanics gives us the information about the operators commutation whether they can be applied in any sequence or not. If the bracket gives zero result then the order will not matter, but if it is non zero then we need to take care of the sequence. The commutator for two operators $\hat A$  and $\hat B$ is defined as,

$[\hat A\hat B]=\hat A\hat B-\hat B\hat A$

Explanation:

Given,

$[x, p] = iℏ$

Hence,

$[x^3, p]=[x\cdot x^2,p]=[x,p]\cdot x^2+x[x^2,p]\ [x^3, p]=i ℏ\cdot x^2+x[x\cdot x,p]=i ℏx^2+x([x,p]\cdot x+x[x,p])\ [x^3, p]=i ℏx^2+i ℏx^2+i ℏx^2=3i ℏx^2$

Hence, the correct answer is option 4.

Concept:

The Schrödinger equation describes how the quantum state of any physical system evolves over space and time. There are two forms of the Schrödinger equation: the time-dependent Schrödinger equation and the time-independent Schrödinger equation.

The time-independent Schrödinger equation represents the conservation of total energy in a quantum system. The Schrödinger equation in its time-independent form is given by:

 $\hat H ψ =Eψ$

where ($\hat{H}$) is the Hamiltonian operator representing the total energy (kinetic + potential) of the system, ($ψ$) is the wave function of the system, and (E) is the energy eigenvalue. The Hamiltonian operator, ($\hat{H}$), usually comprises two parts: the kinetic energy operator, ($\hat{T}$), and the potential energy operator, ($\hat{V}$), such that:

$Hψ=(\hat T +V)ψ$

The Hamiltonian is often expressed as:

$\hat{H} = -\dfrac{\hbar^2}{2m}\nabla^2 + V(\vec{r} )$

where ($\dfrac{\hbar^2}{2m}\nabla^2$) is the kinetic energy term and ($V(\vec{r})$) is the potential energy term. The Schrödinger equation effectively states that the total energy of the system (kinetic + potential) is conserved and is equivalent to the eigenvalue (E) when acting on the wave function (ψ).

Explanation:

Given,

The time-independent Schrödinger equation:

$\hat H \psi=E\psi$

implies that the Hamiltonian operator ($\hat{H}$), which represents the total energy of the system, remains constant over time when applied to the wave function (ψ). It that the total energy of the system is conserved. The eigenvalue (E) corresponds to the measurable energy levels of the system, and the wave function (ψ) describes the probability distribution of the particle's position and momentum.

Hence, the correct answer is option 4.

Concept:

In quantum mechanics, an operator ($\hat{A}$) acting on a wave function (ψ) gives another wave function. An eigenfunction (ψ) of an operator ($\hat{A}$) satisfies the eigenvalue equation:

$\hat Aψ=λ ψ$

where (λ) is the eigenvalue corresponding to the eigenfunction ψ. In this problem, we are given that the operator $\hat A$ acts on the wave functions (ψ₁) and (ψ₂) as follows:

$\hat{A} ψ_1 = ψ_2 \ \hat{A} ψ_2 = ψ_1$

We need to determine the eigenfunctions of the operator (\hat{A}^2).

Explanation:

Given the action of  $\hat{A}$ on ψ1 and ψ2, we apply the operator $\hat{A}$ twice to see how $\hat{A}^2$) acts on these functions.

First, consider ψ1:

$\hat{A}^2 ψ_1 = \hat{A}(\hat{A} ψ_1) = \hat{A} ψ_2 = ψ_1$

Next, consider ψ2:

$\hat{A}^2 ψ_2 = \hat{A}(\hat{A} ψ_2) = \hat{A} ψ_1 = ψ_2$

Hence,

$\hat{A}^2 ψ_1 = ψ_1 \ \hat{A}^2 ψ_2 = ψ_2$

Therefore, both ψ1 and ψ2 are eigenfunctions of the operator $\hat{A}^2$  with eigenvalue 1.

Hence, the correct answer is option 1.

Concept:

The probability density function for a particle in a quantum state is given by the square of the wave function's amplitude, $|\psi(x)|^2$. For a particle in a one-dimensional box (infinite potential well) of length ( L ), the wave functions for the energy states are given by:

$\psi_n(x) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L})$

where, n is the quantum number representing the energy level. For the first excited state, (n = 2). The probability density function for the first excited state is:

$|\psi_2(x)|^2 = \left(\sqrt{\dfrac{2}{L}} \sin\left(\dfrac{2\pi x}{L}\right)\right)^2 = \dfrac{2}{L} \sin^2\left(\dfrac{2\pi x}{L}\right)$

To find the position where the probability density is maximum, we need to find the maximum value of $\sin^2\left(\dfrac{2\pi x}{L}\right)$

Explanation:

For the first excited state n = 2, the wave function is:

$\psi_2(x) = \sqrt{\frac{2}{L}} \sin(\frac{2\pi x}{L})$

The probability density function is:

$|(\psi_2(x)|^2 = \dfrac{2}{L} \sin^2\left(\dfrac{2\pi x}{L}\right)$

The function $\sin^2\left(\dfrac{2\pi x}{L}\right)$ reaches its maximum value of 1.

 $\sin\left(\dfrac{2\pi x}{L}\right) = \pm 1$

Thus, we need to find x such that:

$\dfrac{2\pi x}{L} = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \ldots$

Solving for x, we have:

$x = \dfrac{L}{4}, \dfrac{3L}{4}, \dfrac{5L}{4}, \ldots$

Hence, the function $\sin^2\left(\dfrac{2\pi x}{L}\right)$ is maximum at $x = \dfrac{L}{4} \ \ \text{and} \ \ x = \dfrac{3L}{4}$ within the interval [0, L].

Therefore, the probability of finding the particle in the first excited state is maximum at:

$x = \dfrac{L}{4} \ \ \text{and} \ \ x = \dfrac{3L}{4}$

The correct answer is option 4.

Concept:

The terms C_p and C_v refer to the specific heat capacities of gases at constant pressure and constant volume, respectively. For an ideal gas, C_p is the amount of heat required to raise the temperature of one mole of gas by one degree Celsius at constant pressure, and C_v is the amount of heat required to raise the temperature of one mole of gas by one degree Celsius at constant volume. The difference between these two specific heats for an ideal gas is given by:

\(\text{C}\text{p}-\text{C}\text{v}=\text{R}\)

where, R is the universal gas constant. This relationship is known as Mayer's relation. It indicates that the difference between the molar specific heat at constant pressure and the molar specific heat at constant volume is the same for all ideal gases, and equals the universal gas constant R.

Explanation:

Given,

The specific heat capacities:

Cp for a gas at constant pressure

Cv for a gas at constant volume

Using Mayer^'s relation:

For nitrogen $\text{N}_2$ :

\(\text{x} = \text{C}{\text{p}{\text{N}2}} - \text{C}{\text{v}{\text{N}_2}} = \text{R} \)

For argon $\text{Ar}$:

\(\text{y} = \text{C}{\text{p}{\text{Ar}}} - \text{C}{\text{v}{\text{Ar}}} = \text{R}\)

Since, the gas constant R is a universal constant and has the same value for both nitrogen and argon, we have:

$\text{x} = \text{y} = \text{R}$

Therefore, the relationship between x and y is:

Hence, the correct answer is option 1.

Concept:

In thermodynamics, the efficiency of an engine is a measure of how well it converts heat energy into work.

For a reversible engine operating between two temperatures, the efficiency is given by the Carnot efficiency, which depends only on the temperatures of the heat reservoirs and is the maximum possible efficiency for any heat engine operating between those two temperatures. The Carnot efficiency is given by:

$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$

where, T_H is the temperature of the hot reservoir and T_C is the temperature of the cold reservoir (both in absolute temperature)

For an irreversible engine operating between the same two temperatures, its efficiency is always less than that of a reversible engine. This is because irreversible processes involve additional factors such as friction, unrestrained expansion, and other dissipative effects, which reduce the efficiency.

Explanation:

Given,

Two engines (one reversible and one irreversible) are operating between the same two temperatures T_H and T_C. The efficiency of a reversible engine operating between these two temperatures is given by the Carnot efficiency:

$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$

For the irreversible engine, let's denote its efficiency as $\eta_{\text{irr}}$. The efficiency of an irreversible engine is always less than that of a reversible engine due to the presence of irreversibilities. Hence, we have:

$\eta_{\text{irr}} < \eta_{\text{rev}}$

Therefore, irrespective of the specific details of the irreversible engine, its efficiency will always be less than the Carnot efficiency for the given temperature limits.

Hence, the correct answer is option 4.

Concept:

An ideal black body absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It is a perfect absorber (and emitter) of radiation. No radiation is reflected or transmitted. In addition to being an ideal absorber, an ideal black body is also a perfect emitter of radiation, emitting the maximum possible amount of radiation at a given temperature, following the Planck radiation law.

Physically, no real object behaves as a perfect black body, but certain materials and specific conditions can approximate its behavior. The concept of an ideal black body is significant in thermodynamics and quantum mechanics, particularly in the study of black body radiation, Stefan-Boltzmann law, Wien's displacement law, and Planck's law.

Explanation:

While a perfect black body does not exist, substances such as lamp black, carbon nanotubes, and special cavity designs can closely approximate black body behaviour over certain spectral ranges and conditions.

We know that,

• A perfect absorber of all incident radiation, reflecting and transmitting none.
• A perfect emitter, radiating the maximum possible energy at a given temperature according to Planck's law.

Hence, a hollow enclosure blackened inside will behave as a black body.

The correct answer is option 4.

Concept:

In thermodynamics, when a gas is suddenly compressed by a piston, the process is typically adiabatic, meaning no heat is exchanged with the surroundings during the compression.

This sudden compression results in an increase in the gas's temperature and pressure due to the work done on the gas. When the piston is then maintained at the same position (i.e., the volume is kept constant), the initial state immediately after compression is one of increased pressure and temperature.

Over time, the gas inside the cylinder will gradually exchange heat with its surroundings. This process is described as heat transfer in an isochoric process (constant volume).

Explanation:

Immediately after the sudden compression, the pressure inside the cylinder is high due to the adiabatic compression. However, as time passes, the gas will start to lose heat to the surroundings because the compressed gas is at a higher temperature than the ambient environment.

As the gas loses heat, its internal energy decreases, and since the volume is constant, this results in a decrease in pressure according to the ideal gas law ( PV = nRT ). For constant volume(V):

$P \propto T$

As T decreases, P decreases.

Therefore, over time, the pressure of the gas in the cylinder will decrease as it cools down by transferring heat to the surroundings.

Hence, the correct answer is option 4.

Concept:

In thermodynamics and kinetic theory, the Van der Waals equation provides a realistic description of the behaviour of real gases by accounting for the volume occupied by gas molecules and intermolecular forces. The Van der Waals equation is:

$\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$

where, P is the pressure, V_m is the molar volume, T is the temperature, and (R) is the universal gas constant. The parameters a and b are specific to each gas and account for the attractive forces between molecules and the finite volume occupied by gas molecules, respectively.

For a gas, the critical temperature (T_c), critical pressure (P_c), and critical volume (V_c) are the values at the critical point where the gas and liquid phases become indistinguishable. These critical constants are related to the Van der Waals constants a and b through the following relationships:

$T_c = \frac{8a}{27Rb} \ P_c = \frac{a}{27b^2} \ V_c = 3b$

Explanation:

We are interested in the critical coefficient, defined as:

$\frac{R T_c}{P_c V_c}$

To determine the critical coefficient this we need to substitute the critical constants T_c, P_c, and V_c into the expression.

Given the critical temperature, pressure, and volume relationships:

$T_c = \frac{8a}{27Rb} \ P_c = \frac{a}{27b^2} \ V_c = 3b$