Wavelength of a proton or a neutron, λ ≈ 10^-15 m

Rest mass energy of an electron:

moc² = 0.511 MeV = 0.511 x 10⁶x 1.6 x 10⁻¹⁹ = 0.8176 x 10⁻¹³ J

Planck’s constant, h = 6.6 x 10⁻³⁴Js

Speed of light, c = 3x 10^s m/s

The momentum of a proton or a neutron is given as:

p=h/λ = (6.6 x 10⁻³⁴)/ 10^-15

= 6.6 x 10⁻¹⁹   kg m/s

The relativistic relation for energy (E) is given as:

E = p²c² + m²oc⁴

= (6.6 x 10⁻¹⁹ x 3 x 10⁸ )²+ (0.8176 x 10⁻¹³)² = 392.04 x 10⁻²² + 0.6685 x 10⁻²⁶

≈392.04 x 10⁻²²

Therefore ,E = 1.98 x 10⁻¹⁰

= ( 1.98 x 10⁻¹⁰ )/(1.6 x 10⁻¹⁹) = 1.24 x 10⁹ eV

= 1.24 BeV

Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.

Ans (a).

De Broglie wavelength = 2.327 x 10^-12 m; neutron is not suitable for the diffraction experiment

Kinetic energy of the neutron, K = 150 eV = 150 x 1.6 x 10^-19 = 2.4 x 10^-17 J

Mass of a neutron ,m_n=1.675 x 10^-27 kg

The kinetic energy of the neutron is given by the relation:

K = 1/2 m_nv²

=> m_nv = √(2Km_n)

Where,v = Velocity of the neutron

m_nv = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

λ =h/m_nv   = h/√(2Km_n)

It is clear that wavelength is inversely proportional to the square root of mass.

Hence. wavelength decreases with increase in mass and vice versa.

λ = (6.6 x 10⁻³⁴)/√ [2(2.4 x 10^-17)(1.675 x 10^-27)]

= 2.327 x 10⁻¹² m

It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Aº, i.e., 10^-10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

Ans (b).

De Broglie wavelength = 1.447 x 10^-10 m

Room temperature, T = 27°C = 27 + 273 = 300 K

The average kinetic energy of the neutron is given as:

E= 3/2 kT

Where, k = Boltzmann constant = 1.38 x 10⁻²³ J mol^-1 K⁻¹ The wavelength of the neutron is given as:

λ = h/√(2Em_n) = h/√(3Km_nT )

=   (6.6 x 10⁻³⁴)/√[3 x  (1.675 x  10⁻²⁷ )x (1.38 x  10⁻²³ )x 300]

= 1.447 x 10⁻¹⁰ m

This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high energy neutron beam should first be thermalised, before using it for diffraction.

Intensity of incident light, I = 10^-5 W m⁻²

Surface area of a sodium photocell, A = 2 cm² = 2 x 10^-4 m²

and incident power of the light, P = I x A = 10⁻⁵ x 2 x 10^-4 = 2 x 10^-9 W

Work function of the metal, φ_O = 2 eV = 2 x 1.6 x 10^-19 = 3.2 x 10^–19 J

Number of layers of sodium that absorbs the incident energy, n = 5

We know that the effective atomic area of a sodium atom, A_e is 10^-20 m².

Hence, the number of conduction electrons in n layers is given as:

n’ = n x A/Ae = 5 x (2 x 10^-4) /(10^-20) = 10¹⁷

The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:

E = P/n’

= 5 x (2 x 10^-9)/(10¹⁷) = 2 x 10⁻²⁶ J/s

Time required for photoelectric emission:

t = φ_O/E = (3.2 x 10^–19) /( 2 x 10⁻²⁶)

=1.6 x 10⁷ s ≈ 0.507 years

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.

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.