Q146 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
The ratio of the accelerating potentials required to accelerate (i) an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is
(mass of $\alpha$ - particle = $6.4 \times 10^{-27}$ kg, mass of proton = $1.6 \times 10^{-27}$ kg)
(mass of $\alpha$ - particle = $6.4 \times 10^{-27}$ kg, mass of proton = $1.6 \times 10^{-27}$ kg)
- Correct answerA$1 : 8$
- B$1 : 4$
- C$8 : 1$
- D$4 : 1$
Solution
The de Broglie wavelength is given by $\lambda = \dfrac{h}{\sqrt{2mqV}}$.
Since the de Broglie wavelengths are the same, $\lambda_{\alpha} = \lambda_{p}$.
$\dfrac{h}{\sqrt{2m_{\alpha}q_{\alpha}V_{\alpha}}} = \dfrac{h}{\sqrt{2m_{p}q_{p}V_{p}}}$
$m_{\alpha}q_{\alpha}V_{\alpha} = m_{p}q_{p}V_{p}$
$\dfrac{V_{\alpha}}{V_{p}} = \dfrac{m_{p}q_{p}}{m_{\alpha}q_{\alpha}}$
Given $m_{\alpha} = 6.4 \times 10^{-27}$ kg and $m_{p} = 1.6 \times 10^{-27}$ kg, we have $m_{\alpha} = 4m_{p}$.
The charge of an $\alpha$-particle is twice that of a proton, so $q_{\alpha} = 2q_{p}$.
$\dfrac{V_{\alpha}}{V_{p}} = \dfrac{m_{p} \times q_{p}}{4m_{p} \times 2q_{p}} = \dfrac{1}{8}$
The ratio is $1 : 8$.
Answer: $1 : 8$
Since the de Broglie wavelengths are the same, $\lambda_{\alpha} = \lambda_{p}$.
$\dfrac{h}{\sqrt{2m_{\alpha}q_{\alpha}V_{\alpha}}} = \dfrac{h}{\sqrt{2m_{p}q_{p}V_{p}}}$
$m_{\alpha}q_{\alpha}V_{\alpha} = m_{p}q_{p}V_{p}$
$\dfrac{V_{\alpha}}{V_{p}} = \dfrac{m_{p}q_{p}}{m_{\alpha}q_{\alpha}}$
Given $m_{\alpha} = 6.4 \times 10^{-27}$ kg and $m_{p} = 1.6 \times 10^{-27}$ kg, we have $m_{\alpha} = 4m_{p}$.
The charge of an $\alpha$-particle is twice that of a proton, so $q_{\alpha} = 2q_{p}$.
$\dfrac{V_{\alpha}}{V_{p}} = \dfrac{m_{p} \times q_{p}}{4m_{p} \times 2q_{p}} = \dfrac{1}{8}$
The ratio is $1 : 8$.
Answer: $1 : 8$
Hence, the correct answer is option: 'A'.