Question

Question 2.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1, calculate the energy with which it is bound to the nucleus.

Class XI Structure of Atom Page 68

Answer 1

we know, E = hv = hc/λ
here, h = 6.626 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 150 pm = 1.5 × 10^-10 m

E = 6.626 × 10^-34 × 3 × 10^8/1.5 × 10^-10 J
= 13.25 × 10^-16 J

K.E of ejected electron = 1/2 mv²
= 1/2 × 9.11 × 10^-31 × (1.5 × 10^7)² J
= 1.015 × 10^-16 J

now, energy with which the electron was bound to the nucleus = work function for the metal = w₀

W₀ = hv - 1/2 mv²
= Energy - K.E of ejected electron
= 13.25 × 10^-16 J - 1.025 × 10^-16 J
= 12.225 × 10^-16 J

hence, energy with which the electron was bound to the nucleus = 12.225 × 10^-16 J

Answer 2

Answer:

we know, E = hv = hc/λ

here, h = 6.626 × 10^-34 Js

c = 3 × 10^8 m/s

λ = 150 pm = 1.5 × 10^-10 m

E = 6.626 × 10^-34 × 3 × 10^8/1.5 × 10^-10 J

= 13.25 × 10^-16 J

K.E of ejected electron = 1/2 mv²

= 1/2 × 9.11 × 10^-31 × (1.5 × 10^7)² J

= 1.015 × 10^-16 J

now, energy with which the electron was bound to the nucleus = work function for the metal = w₀

W₀ = hv - 1/2 mv²

= Energy - K.E of ejected electron

= 13.25 × 10^-16 J - 1.025 × 10^-16 J

= 12.225 × 10^-16 J

hence, energy with which the electron was bound to the nucleus = 12.225 × 10^-16 J

Hope it will be helpful!!