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Calculation:

The motion of the electron is in a region where both electric and magnetic fields are acting on it. For the electron to move with constant velocity, the net force acting on it must be zero. The force due to the electric field (F_e) and the magnetic field (F_m) must balance each other.

The force due to the electric field is given by:

F_e = eE

Where 'e' is the charge of the electron and 'E' is the electric field strength.

The force due to the magnetic field is given by:

F_m = evB

Where 'v' is the velocity of the electron and 'B' is the magnetic field strength.

Since the forces must cancel each other out for constant velocity, we set F_e = F_m:

eE = evB

Simplifying, we get:

v = E / B

Substitute the given values:

v = 20 V/m / 0.5 T

v = 40 m/s

The velocity of the electrons must be 40 m/s.

Calculation:

The magnetic field inside a solenoid is given by the formula:

B = μ₀ × n × I

Where:

• μ₀ = 4π × 10^-7 weber amp^-1 m^-1 (permeability of free space)
• n = number of turns per unit length = 10 turns per cm = 10 × 10² turns/m
• I = current passing through the solenoid = 5A

Now, substituting the values into the equation:

B = (4π × 10^-7) × (10 × 10²) × 5

B = 2π × 10^-5 tesla

The magnetic induction at the axis inside the solenoid is 2π × 10^-5 tesla.

Calcultion:

The formula for the induced e.m.f. in a coil is given by:

e = -L × (ΔI / Δt)

Where:

• e = induced e.m.f. = 8 V
• L = self-inductance of the coil (the quantity we need to find)
• ΔI = change in current = 4 A - 2 A = 2 A
• Δt = time taken for the change in current = 0.05 s

Now, substituting the known values into the equation:

8 = L × (2 / 0.05)

8 = L × 40

L = 8 / 40 = 0.2 henry

The coefficient of self-inductance of the coil is 0.2 henry.

Calculation:

The formula for the quality factor (Q) of an LC circuit is given by:

Q = (1 / R) × √(L / C)

Where:

• R = resistance = 100 Ω
• L = inductance = 80 mH = 80 × 10^-3 H
• C = capacitance = 2 μF = 2 × 10^-6 F

Substituting the values into the formula:

Q = (1 / 100) × √((80 × 10^-3) / (2 × 10^-6))

Q = (1 / 100) × √(40 × 10³)

Q = (1 / 100) × 200

Q = 2

The quality factor of the circuit is 2.

Calculation:

The pressure exerted by an electromagnetic wave on a surface can be calculated by the formula:

Pressure (P) = Energy / (Area × Time)

We are given:

• Energy = E
• Area = A
• Time = t
• Speed of light = c

Since the wave is perfectly reflected, the pressure is twice that of a non-reflected wave.

Therefore, the formula for the average pressure exerted on the surface is:

P = 2E / (A × c)

The correct answer is: 2E / (A × c)

Calculation:

The fringe separation in a Young's double slit experiment is given by the formula:

y = (λ × D) / d

Where:

• λ is the wavelength of light
• D is the distance between the slits and the screen
• d is the distance between the slits

Substitute the given values:

• λ = 500 nm = 500 × 10^-9 m
• D = 1 m
• d = 2 mm = 2 × 10^-3 m

The fringe separation (y) is:

y = (500 × 10^-9 × 1) / (2 × 10^-3) = 0.25 mm

The correct answer is: 0.25 mm

Calculation:

The number of waves in a medium is given by:

n = (thickness of medium) / (wavelength in that medium)

For the same number of waves to exist in both the water and the glass slab, the relationship between the number of waves in water and in glass can be expressed as:

Number of waves in water = Number of waves in glass

This can be written as:

(thickness of water) / (wavelength in water) = (thickness of glass) / (wavelength in glass)

Now, using the refractive index (n) to relate the wavelengths:

wavelength in medium = (wavelength in vacuum) / refractive index of medium

Let the refractive index of glass be "n_g" and the refractive index of water is given as 4/3. Substituting the values:

(5 cm) / (wavelength in water) = (4 cm) / (wavelength in glass)

Using the refractive index formula:

(5 cm) / (wavelength in vacuum / (4/3)) = (4 cm) / (wavelength in vacuum / n_g)

Solving for n_g:

n_g = (5/4) × (4/3) = 5/3

The correct answer is: 5/3

Calculation:

The de Broglie wavelength λ is given by the formula:

λ = h / p

where:

• h = Planck's constant
• p = momentum of the particle

Since the steel ball is moving with kinetic energy K, the momentum p is related to K by the equation:

p = √(2mK)

Substitute this expression for p into the de Broglie wavelength equation:

λ = h / √(2mK)

The correct answer is: h / √(2mK)

Calculation:

Let the work function of the metal be φ, and let the initial wavelength be λ₁ = 500 nm and the final wavelength be λ₂ = 200 nm. The energy of the photon is given by:

E = h× c / λ

where:

• h = Planck's constant
• c = speed of light
• λ = wavelength

Initially, the kinetic energy of the photoelectron is:

K.E. = E1 - φ = h × c / λ1 - φ

Finally, the kinetic energy becomes three times larger:

3 × K.E. = E2 - φ = h × c / λ2 - φ

From the ratio of the kinetic energies:

3 × (h × c / λ1 - φ) = h × c / λ2 - φ

Solve for φ (work function of the metal):

φ ≈ 0.61 eV

The correct answer is: 0.61 eV

Calculation:

The given figure represents a p-type semiconductor:

When one of the silicon atoms (with valence = 4) is replaced by an atom of aluminium (with valence = 3), the aluminium atom can bond covalently with only three silicon atoms. This creates a "missing" electron (a hole) in one of the aluminium-silicon bonds.

With a small expenditure of energy, an electron can be torn from a neighboring silicon-silicon bond to fill this hole, thereby creating a hole in that bond as well. This process continues as the hole migrates through the lattice.

The aluminium atom is known as an acceptor atom because it readily accepts an electron from a neighboring bond in the valence band of silicon.

As the figure suggests, this electron occupies a localized acceptor state that lies within the energy gap, at an average energy interval E_a above the top of the valence band.

By adding acceptor atoms, it is possible to significantly increase the number of holes in the valence band, thereby making the material a p-type semiconductor.

The material shown in the figure is a p-type semiconductor.