Wavelength of the monochromatic radiation,

λ = 640.2 nm = 640.2 x 10⁻⁹ m

Stopping potential of the neon lamp, Vo = 0.54 V

Charge on an electron, e = 1.6 x 10⁻¹⁹ C

Planck’s constant, h = 6.6 x 10^-34 Js

Let φ_O be the work function and v be the frequency of emitted light.

We have the photo-energy relation from the photoelectric effect as: eV₀ = hv — φ_O,

φ_O = hc/λ  – eV₀

₌ ( 6.6 x 10^-34) x ( 3 x 10⁸ ) /( 640.2 x 10⁻⁹) – (1.6 x 10⁻¹⁹) x 0.54

= 3.093 x 10⁻¹⁹ -0.864 x 10⁻¹⁹ = 2.229 x10⁻¹⁹ J

=   (2.229 x10⁻¹⁹) /(1.6 x 10⁻¹⁹) = 1.39 eV

Wavelength of the radiation emitted from an iron source,

λ’ = 427.2 nm = 427.2 x 10^-9 m

Let V₀ be the new stopping potential. Hence, photo-energy is given as:

eV0′ = hc/λ’- φ_O

=(6.6 x 10^-34) x (3 x 10⁸ ) /(427.2 x 10^-9) – (2.229 x10⁻¹⁹)n

= 4.63 x 10⁻¹⁹ – 2.229 x10⁻¹⁹

=2.401 x 10⁻¹⁹J

= (2.401 x 10⁻¹⁹)/(1.6 x x10⁻¹⁹) = 1.5eV

Hence, the new stopping potential is 1.50 eV.

Ans (a) .

Power of the medium wave transmitter, P = 10 kW = 10⁴W = 10⁴ J/s

Hence, energy emitted by the transmitter per second, E = 10⁴ and

wavelength of the radio wave, λ= 500 m

The energy of the wave is given as:

E₁ = hc/λ

Where, h = Planck’s constant = 6.6 x 10⁻³⁴ Js and c = Speed of light = 3 x 10⁸ m/s

Therefore, E₁ = (6.6 x 10⁻³⁴) x (3 x 10⁸) /(500) = 3.96 x 10⁻²⁸  J

Let n be the number of photons emitted by the transmitter.

Therefore , nE₁ = E

n = E/E₁

= 10⁴/(3.96 x 10⁻²⁸ ) =  2.525 x 10³¹

≈ 3 x 10³¹

The energy (E1) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large.

The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.

Ans (b).

Intensity of light perceived by the human eye, I = 10⁻¹⁰ W m^-2

Area of a pupil, A = 0.4 cm² = 0.4 * 10⁻⁴ m²

Frequency of white light, v= 6 x 10¹⁴ Hz

The energy emitted by a photon is given as:

E = hv Where,

h = Planck’s constant = 6.6 x 10^-34 Js

Therefore, E = 6.6 x 10-³⁴ x 6 x 10¹⁴ = 3.96 x 10⁻¹⁹ J

Let n be the total number of photons falling per second, per unit area of the pupil.

The total energy per unit for n falling photons is given as:

E = n x 3.96 x 10^-19 J s⁻¹ m⁻²

The energy per unit area per second is the intensity of light.

Therefore, E = I

n x 3.96 x 10^–19 = 10^–10

n  =10^–10/3.96 x 10^–19

= 2.52 x 10⁸ m² s^–1

The total number of photons entering the pupil per second is given as:

n_A = n x A = 2.52 x 10⁸ x 0.4 x 10^–4 = 1.008 x 10⁴ s^–1

This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.