Question

An X-ray photon with a wavelength of 0.979 nm strikes a surface. The emitted electron has a kinetic energy of 984 eV.

What is the binding energy of the electron in kJ/mol? [Note that KE = 12mv2 and 1 electron volt (eV) = 1.602×10−19J.]

Answer 1

K.E of electron = 984 eV

We have relation, 1 eV = 1.602 10⁻¹⁹ J

K.E of electron = 984 eV ( 1.602 10⁻¹⁹ J / 1 eV ) = 1.576 10⁻¹⁶ J

We have relation, h = h₀ + 1/2 mv ² ---------------------> (1)

where, h is the energy of incident photon of light, h₀ is binding energy of the electron and 1/2 mv ² is a K.E of electron.

First calculate energy of incident photon of light.

We have relation, 1 nm = 10 ^-09 m

Wavelength of incident light = 0.979 nm   ( 10 ^-09 m / 1 nm )

Wavelength of incident light = 9.79 10 ^-10 m

We have, Energy  = h C /

Where, h is a planck constant , C is velocity of light and is wavelength of light.

Energy of incident light = ( 6.626 10 ^{- 34} J . s ) ( 3.00 10 ⁸ m /s ) / 9.79 10 ^-10 m

Energy of incident light = 2.030 10 ^{- 16} J

Substituting energy of incident light = 2.030 10 ^{- 16} J and K.E = 1.576 10⁻¹⁶ J in relation 1 , we get

2.030 10 ^{- 16} J =   h₀ + 1.576 10⁻¹⁶ J

h₀ = 2.030 10 ^{- 16} J - 1.576 10⁻¹⁶ J

h₀ = 4.54 10⁻¹⁷ J

ANSWER: Binding energy of electron = h₀ = 4.54 10⁻¹⁷ J