Question

Find the de Broglle wavelength of a 40 eV electron

Answer 1

Answer:

Hi

Explanation:

1. The energy of the electron is given as E = 40 keV.

2. The de Broglie wavelength 'lambda' is defined as.

3. lambda = h / p.

4. where h is Planck 's constant and p is the momentum of the particle.

5. we know E = p^2 / 2.m.

6. therefore p = Sqrt( 2.m.E)

7. lambda = h / Sqrt( 2.m.E) = 4.1357 × 10-15 eV s / {2. ...

8. lambda = 7 x 10^-12 m approx.

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