Consider an electron accelerated from rest through a potential difference V. Let v be the final speed of the electron. We consider the nonrelativistic case, v << c, where c is the speed of light in free space. The kinetic energy acquired by the electron is

\(\cfrac12\) mv² = \(\cfrac1{2m}\)(mv)² = eV......(1)

where e and m are the electronic charge and mass (nonrelativistic).

Therefore, the electron momentum,

p = mv = \(\sqrt{2meV}\).....(2)

The de Broglie wavelength associated with the electron is

λ = \(\cfrac hp\)....(3)

where h is Planck’s constant. 

From Eqs. (2) and (3),

λ = \(\cfrac h{\sqrt{2meV}}\)

Equation (4) gives the required expression. 

[Note : Substituting the values of h = 6.63 × 10^-34 J-s, 

m = 9.1 × 10^-31 kg, 

e = 1.6 × 10^-19 C in 

Eq. (4), we obtain 

λ = \(\sqrt{150/V}\) + 10^-10 m = \(\sqrt{150/V}\) Å 

= 12.25 / √V Å, where V is in volt. 

Therefore, electrons accelerated from rest through 150 volts have a de Broglie wavelength of 1 Å. This corresponds to the Xray region of the electromagnetic spectrum.]