Question

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 $1.5 \times 10^{7}$ m$s^{-1}$, calculate the energy with which it is bound to the nucleus.

Answer 1

"Energy of incident photon (E) is given by,

$E\quad =\quad \frac { hc }{\lambda}$

$=\quad \frac { \left( 6.626\quad \times \quad { 10 }^{ -34 }\quad Js \right) \left( 3.0\quad \times \quad { 10 }^{ 8 }\quad m/s \right)}{\left( 150\quad \times \quad { 10}^{ -12}\quad m \right)}$

$=\quad 1.3252\quad \times \quad { 10 }^{ -15 }\quad J$

$E\quad \simeq \quad 13.252\quad \times \quad { 10 }^{ -16 }\quad J$

Energy of the electron ejected (K.E)

$=\quad \frac { 1 }{ 2 } { m }_{ e }{ v }^{ 2 }$

$=\quad \frac { 1 }{ 2 } \left( 9.10939\quad \times \quad { 10 }^{ -31 }\quad kg \right) { \left( 1.5\quad \times \quad { 10 }^{ 7 }\quad m/s \right)}^{ 2 }$

$=\quad 10.2480\quad \times \quad { 10 }^{ -17 }\quad J$

$K.E\quad =\quad 1.025\quad \times \quad { 10 }^{ -16 }\quad J$

Energy with which the electron was bound to the nucleus

$=\quad 13.25\quad \times \quad { 10 }^{ -16 }\quad J\quad -\quad 1.025\quad \times \quad { 10 }^{ -16 }\quad J$

$=\quad 12.225\quad \times \quad { 10 }^{ -16 }\quad J$

$=\quad \frac { 12.225\quad \times \quad { 10 }^{ -16 }\quad J }{ 1.602\quad \times \quad { 10 }^{ -19 }\quad eV }$

$E\quad =\quad 7.63\quad \times \quad { 10 }^{ 3 }\quad eV$"

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