JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 1)
The ratio of de-Broglie wavelength of an $$\alpha$$ particle and a proton accelerated from rest by the same potential is $$\frac{1}{\sqrt m}$$, the value of m is -
2
16
8
4
Explanation
Here : $m_\alpha=4 m_P, q_\alpha=2 q_P$, potential $=V$
$\lambda=\frac{h}{\sqrt{2 m q V}}$
So, $\frac{\lambda_\alpha}{\lambda_P}=\sqrt{\frac{2 m_P q_P V}{2 m_\alpha q_\alpha V}}=\sqrt{\frac{m_P \cdot q_P}{4 m_P \times 2 q_P}}=\frac{1}{\sqrt{8}}$
$$ \Rightarrow $$ $m=8$
$\lambda=\frac{h}{\sqrt{2 m q V}}$
So, $\frac{\lambda_\alpha}{\lambda_P}=\sqrt{\frac{2 m_P q_P V}{2 m_\alpha q_\alpha V}}=\sqrt{\frac{m_P \cdot q_P}{4 m_P \times 2 q_P}}=\frac{1}{\sqrt{8}}$
$$ \Rightarrow $$ $m=8$
Comments (0)
