Question

An X-ray photon of wavelength 0.989 nm strikes the surface of a material. The emitted photoelectron has a speed of 1.85 × 107 m s–1 . What is the binding energy of the electron in a.) eV and b.) kJ/mol?

Answer #1

**Kinetic energy of electron KE = 1/2 x m x
v^2**

**
= 0.5 x 9.1 x 10^-31 kg x (1.85 x 10^7 ms-1)^2 = 1.56 x 10^-16
kgm2s-2 = 1.56 x 10^-16 J**

**energy of incident photon = h c / lmada = ( 6.625 x
10^-34 Js x 3 x 10^8 m/s) / ( 0.989 x 10^ -9 m)**

** = 2 x 10^-16
J**

**a) Binding energy = incidnet energy of photon - kinectic
energy of ejected electron**

**
= ( 2 x 10^-16 J) - ( 1.56 x 10^-16 J) = 4.4 x 10^-17
J**

**Binding energy in eV = ( 4.4 x 10^-17 J) / ( 1.602 x
10^-19 J) = 277 eV**

**b) Biinding energy per mole = 4.44 x 10^-17 J x 6.023 x
10^23 = 2.65 x 10^7 J**

**BE in KJ/mol = 2.65 x 10^7 x 10^-3 KJ = 2.65 x 10^4
KJ/mol = 26501 KJ/mol**

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Part A
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