Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Options

A. 102 pm
B. 124 pm
C. 95 pm
D. 112 pm

Answer

D

Explanation

The de - Brogile wavelength of an electron is
` lambda = h/(sqrt(2mK))`
where h is the Planck's constant , m is mass of the electron and K is its Kinetcic energy .
Here ,
`h = 6.63 xx10^(-34)J s `
` m = 9.1 xx10^(-31) kg`
`K = 120 eV = 120 xx 1.6 xx10^(19) J `
`( :' 2 eV = 1.6 xx10^(-19) J )`
` :. lambda = (6.63 xx10^(-34) J s )/(sqrt(2(9.1 xx10^(-31)kg)(120 xx1.6 xx10^(-19)J )))`
` = 1.12 xx10^(-10) m = 112 xx 10^(-12) m `
112 pm `( :' 1 "pm" = 10^(-12) m )`