Question

Electrons are accelerated through a potential difference $V_e$, and protons are accelerated through a potential difference of 4 V. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons, respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by (given $m_e$ is mass of electrons and $m_p$ is mass of protons):

• A. $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_e}{m_p}}$
• B. $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$
• C. $\frac{\lambda_e}{\lambda_p} = \frac{1}{\sqrt{\frac{m_p}{m_e}}}$
• D. $\frac{\lambda_e}{\lambda_p} = 2 \sqrt{\frac{m_p}{m_e}}$

Hint:

De-Broglie wavelength is inversely proportional to the square root of the mass of the particle. For electrons and protons, their wavelengths depend on their respective masses.

Answer 1

The Correct Option is D

The de-Broglie wavelength of a particle is given by the equation: $$\lambda = \frac{h}{\sqrt{2m eV}}$$ where $h$ is Planck’s constant, $m$ is the mass of the particle, $e$ is the charge of the particle, and $V$ is the potential difference.
For electrons and protons, their respective wavelengths are: $$\lambda_e = \frac{h}{\sqrt{2m_e e V_e}}, \quad \lambda_p = \frac{h}{\sqrt{2m_p e V_p}}$$ Since the potential difference is the same (4 V), the ratio $\frac{\lambda_e}{\lambda_p}$ is: $$\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_p}}{\sqrt{m_e}} = 2 \sqrt{\frac{m_p}{m_e}}$$
Thus, the correct answer is (d).