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Q.
An electron and a proton have same de-Broglie wavelength, then kinetic energy of the electron is:
Bihar CECEBihar CECE 2001Dual Nature of Radiation and Matter
Solution:
de-Broglie wavelength of a charged particle is given by
$\lambda=\frac{h}{m v} \ldots$ (i)
where $h$ is Plancks constant.
If kinetic energy of particle of mass $m$ is $v$, then
$K=\frac{1}{2} m v^{2}$
$\Rightarrow v=\sqrt{\frac{2 K}{m}}$....(ii)
Combining Eqs. (i) and (ii), we get
$\lambda=\frac{h}{m \sqrt{\frac{2 K}{m}}}=\frac{h}{\sqrt{2 m K}} . \ldots$.(iii)
$\therefore \lambda_{e}=\frac{h}{\sqrt{2 m_{e} K_{e}}}$
and $\lambda_{p}=\frac{h}{\sqrt{2 m_{p} K_{p}}}$
but $\lambda_{e}=\lambda_{p}$ (given)
$\because \frac{h}{\sqrt{2 m_{e} K_{e}}}=\frac{h}{\sqrt{2 m_{p} K_{p}}}$
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 $\frac{m_{p}}{m_{e}}>1$
$\therefore \frac{K_{e}}{K_{p}}>1$
$\Rightarrow K_{e} >K_{p}$
Note : If an electron is accelerated through a potential difference of $V$ volt, then Eq. (iii) takes the form $\lambda=\frac{h}{\sqrt{2 m e V}}$
After putting the numerical values for electron, we get
$\lambda=\sqrt{\frac{150}{V}} \mathring{A}$