Question

If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of 1.5 × 10^7 m s–1, calculate the energy with which it is bound to the nucleus.
Ncert solutions for Class 11th Chemistry Part - 1 Chapter 2 Exercise 54

Answer 1

E=6.6*10^-34*3*10^8/1.5*10^-10=1.32*10^-15
Energy of ejected electron 
E=1/2mv2
=1/2*9.1*10^-31*(1.5*10^7)2
=1.02*10^-16

Answer 2

we know, E = hv = hc/λ

here, h = 6.626 × 10⁻³⁴ Js

c = 3 × 10⁸ m/s

λ = 150 pm = 1.5 × 10⁻¹⁰ m

E = 6.626 × 10⁻³⁴ × 3 × 10⁸/1.5 × 10⁻¹⁰ J

= 13.25 × 10⁻¹⁶ J

K.E of ejected electron = 1/2 mv²

= 1/2 × 9.11 × 10⁻³¹ × (1.5 × 10⁷)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 WAS HELPFUL