Question

An electron and a photon have same wavelength. If $p$ is the momentum of electron and $E$ is the energy of photon, the magnitude of $\frac{p}{E}$ in SI unit will be :

• A. $3.0\times {10}^8$
• B. $3.33\times {10}^{-9}$
• C. $9.1\times {10}^{-31}$
• D. $6.64\times {10}^{-34}$

Answer 1

The correct option is B $3.33\times {10}^{-9}$

From de Broglie wavelength formula
$\lambda =\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }....\left(1\right)$
And we know that energy of photon
$E=hv$
$E=\frac{hc}{\lambda }........\left(2\right)$
$\therefore$ From equation (1) and (2),
$\frac{p}{E}=\frac{h}{\lambda }\times \frac{\lambda }{hc}=\frac{1}{c}$
$\frac{p}{E}=\frac{1}{3\times {10}^8}=3.33\times {10}^{-9}$