Question

An electron of velocity v is found to have a certain value of de broglie wavelength. The velocity possessed by neutron to have the same de broglie wavelength is:

Answer 1

The answer to the question will be  $=5.44\times 10^-4\ v$

CALCULATION:

From de Broglie theory, we know that a moving microscopic particle is associated with a wave called matter wave. The wavelength of the particle associated with a particle of mass m moving with velocity v is calculated as -

                                wavelength $\lambda=\frac{h}{mv}$

Here, m is the mass of the body and h is the Planck's constant.

Let us consider the masses of electron and neutron are m and m'.

As per the question, the wavelength associated with electron and neutron is same.

Let the velocity of neutron is v'.

The velocity of electron is given as v.

Hence, the wavelength associated with electron is $\lambda =\frac{h}{mv}$

The wavelength associated with neutron is $\lambda=\frac{h}{m'v'}$

As the wavelength in both the situation are same, so we may write-

                                   $\frac{h}{mv}=\frac{h}{m'v'}$

                                   $m'v'=mv$

                                   $v'=\frac{mv}{m'}$

                                         $=\frac{9.11\times 10^-31v}{1.675\times 10^-27}$

                                         $=5.44\times 10^-4\ v$    [ans]