Question

What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Answer 1

Given: The kinetic energy of the electron is 120 eV.

a)

The momentum of electron is given as,

p= 2m E k

Where, m is the mass of the electron and E k is the kinetic energy of electron.

By substituting the given values in the above equation, we get

p= 2×9.1× 10 −31 ×( 120×1.6× 10 −19 ) =5.91× 10 −24  kg  ms -1

Thus, the momentum of the electron is 5.91× 10 −24  kg⋅ ms −1 .

b)

The speed of the electron is given as,

v= p m

By substituting the given values in the above equation, we get

v= 5.91× 10 −24 9.1× 10 −31 =6.49× 10 6  m/s

Thus, the speed of electron is 6.49× 10 6  m/s.

c)

de Broglie wavelength of electron is given as,

λ= h p

Where, Planck’s constant is h.

By substituting the given values in above equation, we get

λ= 6.626× 10 −34 5.91× 10 −24 =1.121× 10 −10  m =0.112 nm

Thus, the de Broglie wavelength of the electron is 0.112 nm.