Question

Find the ratio of de-Broglie wavelength of a proton and a 𝝰 - particle, when accelerated through a potential difference of 2V and 4V respectively.

• A. 4:1
• B. 2:1
• C. 1:8
• D. 16:1

Answer 1

The Correct Option is A

Solution:
Given:

$V_p = 2V$

$V_\alpha = 4V$

$q_p = e$

$q_\alpha = 2e$

$m_p = m_p$

$m_\alpha = 4m_p$

$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p \cdot 2e \cdot 4V}{m_p \cdot e \cdot 2V}}$
$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{32 m_p e V}{2 m_p e V}}$
$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{16}$
$\frac{\lambda_p}{\lambda_\alpha} = 4$ Therefore, the ratio is $4:1$. The correct answer is (1).