Q111 / 150 · MHT CET 2024 (03 May Shift 1) · Physics
The ratio of the wavelength of a photon of energy ' $E$ ' to that of the electron of same energy is ( $\mathrm{m}=$ mass of an electron, $\mathrm{c}=$ speed of light, $\mathrm{h}=$ Planck's constant)
- A$\sqrt{\frac{\mathrm{m}}{\mathrm{cE}}}$
- B$\sqrt{\frac{2 \mathrm{~m}}{\mathrm{cE}}}$
- C$\mathrm{c} \sqrt{\frac{\mathrm{m}}{\mathrm{E}}}$
- Correct answerD$\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}$
Solution
We know,
Energy of a photon $E=\frac{h c}{\lambda_p}$
$\therefore \quad \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$
Wavelength of an electron $\lambda_e=\frac{h}{m v}$ We also know, $E=\frac{1}{2} m v^2 \Rightarrow v=\sqrt{\frac{2 E}{m}}$ $\begin{array}{ll} \therefore \quad & m v=\sqrt{2 E m} \\ \therefore \quad & \lambda_e=\frac{h}{\sqrt{2 E m}} \\ & \Rightarrow \frac{\lambda_p}{\lambda_e}=\frac{h c}{E} \times \frac{\sqrt{2 E m}}{h}=c \sqrt{\frac{2 m}{E}} \end{array}$
Wavelength of an electron $\lambda_e=\frac{h}{m v}$ We also know, $E=\frac{1}{2} m v^2 \Rightarrow v=\sqrt{\frac{2 E}{m}}$ $\begin{array}{ll} \therefore \quad & m v=\sqrt{2 E m} \\ \therefore \quad & \lambda_e=\frac{h}{\sqrt{2 E m}} \\ & \Rightarrow \frac{\lambda_p}{\lambda_e}=\frac{h c}{E} \times \frac{\sqrt{2 E m}}{h}=c \sqrt{\frac{2 m}{E}} \end{array}$
Hence, the correct answer is option: 'D'.