Question

Question 2.9 A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate

(i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).

Class XI Structure of Atom Page 65

Answer 1

concept :
energy of photons E = hc/λ
where,
h is Plank's constant
c is the speed of light
λ is the wavelength of photons

(i) here, h = 6.626 × 10^-34 Js , c = 3 × 10^8 m/a
and λ = 4 × 10^-7 m .
Now,
E = 6.626 × 10^-34 × 3 × 10^8/4 × 10^-7 J
= 4.969 × 10^-19 J
we know,
1 ev = 1.602 × 10^-19 J
so, E = 4.969 × 10^-19 / 1.602 × 10^-19 ev
= 3.10 ev

(ii) according to Einstein photoelectric theory,
Kinetic energy = hv - hv₀
= 3.10 - 2.13 = 0.97 ev

(iii) Kinetic energy = 1/2mv² = 0.97 ev
1/2 × 9.11 × 10^-31 kg × v² = 0.97 × 1.602 × 10^-19J
v² = 0.97 × 1.602 × 10^-19 × 2 / 9.11 × 10^-31
= 0.341 × 10¹² m²/s²
v = 0.584 × 10^6 m/s

Answer 2

Answer:

Explanation:energy of photon:3.10eV

K. E.of emmision:0.97eV

Velocity :5.84x10^5m/s