An electron and a proton have same de-Broglie wavelength, then kinetic energy of the electron is:

Question

An electron and a proton have same de-Broglie wavelength, then kinetic energy of the electron is: (A) infinite (B) equal to KE of proton (C) zero (D) greater than KE of proton

Tardigrade Admin · Accepted Answer

de-Broglie wavelength of a charged particle is given by

λ = \frac{h}{m v} \dots

(i)
where

h

is Plancks constant.
If kinetic energy of particle of mass

m

is

v

, then

K = \frac{1}{2} m v^{2}

\Rightarrow v = \frac{2 K}{m}

....(ii)
Combining Eqs. (i) and (ii), we get

λ = \frac{h}{m \frac{2 K}{m}} = \frac{h}{2 m K} . \dots

.(iii)

∴ λ_{e} = \frac{h}{2 m _{e} K _{e}}

and

λ_{p} = \frac{h}{2 m _{p} K _{p}}

but

λ_{e} = λ_{p}

(given)

∵ \frac{h}{2 m _{e} K _{e}} = \frac{h}{2 m _{p} K _{p}}

or

2 m_{e} K_{e} = 2 m_{p} K_{p}

or

\frac{K _{e}}{K _{p}} = \frac{m _{p}}{m _{e}}

Since,

m_{p} > m_{e}

or

\frac{m _{p}}{m _{e}} > 1

∴ \frac{K _{e}}{K _{p}} > 1

\Rightarrow K_{e} > K_{p}

Note : If an electron is accelerated through a potential difference of

V

volt, then Eq. (iii) takes the form

λ = \frac{h}{2 m e V}

After putting the numerical values for electron, we get

λ = \frac{150}{V} \overset{˚}{A}

Q. An electron and a proton have same de-Broglie wavelength, then kinetic energy of the electron is:

infinite

equal to KE of proton

zero

greater than KE of proton