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 -
Solution
Here : $m_\alpha=4 m_P, q_\alpha=2 q_P$, potential $=V$
<br/><br/>$\lambda=\frac{h}{\sqrt{2 m q V}}$
<br/><br/>So, $\frac{\lambda_\alpha}{\lambda_P}=\sqrt{\frac{2 m_P q_P V}{2 m_\alpha q_\alpha V}}=\sqrt{\frac{m_P \cdot q_P}{4 m_P \times 2 q_P}}=\frac{1}{\sqrt{8}}$
<br/><br/>$\Rightarrow$ $m=8$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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