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the de broglie wavelength of a photon is twice the de broglie wavelength of an electron

Question :- the de broglie wavelength of a photon is twice the de broglie wavelength of an electron. The speed of the electron is v_e = c/100. Then –

$\displaystyle (A)\text{ }\frac{{{E}_{e}}}{{{E}_{p}}}={{10}^{-4}}$

$\displaystyle (B)\text{ }\frac{{{E}_{e}}}{{{E}_{p}}}={{10}^{-2}}$

$\displaystyle (C)\text{ }\frac{{{p}_{e}}}{{{m}_{e}}c}={{10}^{-2}}$

$\displaystyle (D)\text{ }Both\text{ }(B)\text{ }and\text{ }(C)$

Solution :-

$\displaystyle {{\lambda }_{e}}=\frac{h}{{{p}_{e}}}=\frac{h}{{{m}_{e}}{{v}_{e}}}=\frac{h}{{{m}_{e}}\left( \frac{c}{100} \right)}=\frac{100h}{{{m}_{e}}c}$

According to question, λ_p = 2λ_e, so

$\displaystyle {{\lambda }_{p}}=\frac{200h}{{{m}_{e}}c}$

Energy of photon,

$\displaystyle {{E}_{p}}=\frac{hc}{{{\lambda }_{p}}}=\frac{hc}{\frac{200h}{{{m}_{e}}c}}=\frac{{{m}_{e}}{{c}^{2}}}{200}$

Kinetic energy of electron,

$\displaystyle {{E}_{e}}=\frac{1}{2}{{m}_{e}}v_{e}^{2}=\frac{1}{2}{{m}_{e}}{{\left( \frac{c}{100} \right)}^{2}}=\frac{{{m}_{e}}{{c}^{2}}}{20000}$

Hence,

$\displaystyle \frac{{{E}_{e}}}{{{E}_{p}}}=\frac{{{m}_{e}}{{c}^{2}}}{20000}\times \frac{200}{{{m}_{e}}{{c}^{2}}}={{10}^{-2}}$

This is option (B).

Again,

$\displaystyle \frac{{{p}_{e}}}{{{m}_{e}}c}=\frac{{{m}_{e}}{{v}_{e}}}{{{m}_{e}}c}=\frac{{{v}_{e}}}{c}=\frac{c/100}{c}=\frac{1}{100}={{10}^{-2}}$

This is option (C).

Hence option (D) is correct.

the de broglie wavelength of a photon is twice the de broglie wavelength of an electron

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