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QJEEM23Ph7953: JEE Main 2023 Physics - Class 12 JEEM23Ph7953 Solution: The ratio of de-Broglie wavelength of an particle and a proton…

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 -

Multiple Choice Options

A 2
B 16
C 8
D 4

Step-by-Step Verified Solution

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

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