Question

An electron falls through a potential difference of 100 volt. Calculate the momentum of the electron and the length of the wave associated with the electron in motion. How could these waves be detected..?​

Answer 1

Explanation:

Using energy conservation, KE=PE or

2m

p

2

=eV or p=

2meV

where p= momentum of electron, m= mass of electron, e= charge of electron and V= potential difference.

Now, de-Broglie wavelength λ=

p

h

=

2meV

h

=

2(9.1×10

−31

)(1.6×10

−19

)(100)

6.6×10

−34

=1.22×10

−10

m

Answer 2

Answer:

The de-broglie wavelength of a particle with momentum p is λ=ph

⟹p=λh

Hence the kinetic energy of the particle=2mp2

=2mλ2h2

Hence the kinetic energy of the electron=2meeλe2h2eV

=150eV