An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

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Answer 1

For electron
De - Broglie wavelength $λ_{c} = \frac{h}{p}$
where p is momentum $p = m v$
also by energy we have $E = \frac{1}{2} m v^{2}$

$\Rightarrow E = \frac{1}{2} \frac{p ^{2}}{m}$

$\Rightarrow p = 2 m E$

$∴ λ_{c} = \frac{h}{2 m E}$

For photon energy $\Rightarrow E = \frac{h c}{λ}$

$\Rightarrow λ = \frac{h c}{E}$

$∴ \frac{λ _{c}}{λ} = \frac{h}{2 m E} \frac{E}{h c}$

$= \frac{1}{C} \frac{E}{2 m}$

option (A) is correct.

Answer 2

For electron

De - Broglie wavelength

λ_{c} = \frac{h}{p}

where p is momentum

p = m v

also by energy we have

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

\Rightarrow E = \frac{1}{2} \frac{p ^{2}}{m}

\Rightarrow p = 2 m E

∴ λ_{c} = \frac{h}{2 m E}

For photon energy

\Rightarrow E = \frac{h c}{λ}

\Rightarrow λ = \frac{h c}{E}

∴ \frac{λ _{c}}{λ} = \frac{h}{2 m E} \frac{E}{h c}

= \frac{1}{C} \frac{E}{2 m}

option (A) is correct.