An electron of mass m and a photon have the same energy E in the range of a few eV.


An electron (of mass $m$ ) and a photon have the same energy $E$ in the range of a few $\mathrm{eV}$. The ratio of the deBroglie wavelength associated with the electron and the wavelength of the photon is $(c=$ speed of light in vacuum $)$

  1. \text { (1) } \frac{1}{c}\left(\frac{2 E}{m}\right)^{1 / 2}

  2. (2) $c(2 m E)^{1 / 2}$

  3. (3) $\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}$

  4. (4) $\left(\frac{E}{2 m}\right)^{1 / 2}$

Correct Option: , 3


(3) De-Broglie wavelength of electron $\left(\lambda_{e}\right)$ is given by

$\lambda_{e}=\frac{h}{p_{e}}=\frac{h}{\sqrt{2 m E}}$           $\ldots$ (i) $\quad(\because p=\sqrt{2 m E})$

Energy of photon $(E)$ is given by

$E=\frac{h c}{\lambda_{P}}$        (Here $\lambda p=$ wavelength of Photon)

$\Rightarrow \lambda_{P}=\frac{h c}{E}$ ...(ii)

On dividing (i) by (ii) we get

$\Rightarrow \quad \frac{\lambda_{e}}{\lambda_{P}}=\frac{h}{\sqrt{2 m E}} \frac{E}{h c}=\sqrt{\frac{E}{2 m}} \cdot \frac{1}{c}$



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