Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are \(\lambda_e\) and \(\lambda_p\) for electrons and protons respectively. The ratio of \(\frac{\lambda_e}{\lambda_p}\) is given by: (given \(m_e\) is mass of electron and \(m_p\) is mass of proton).
Show Hint
- \(\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}\)
- \(\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_e}{m_p}}\)
- \(\frac{\lambda_e}{\lambda_p}=\frac{1}{2}\sqrt{\frac{m_e}{m_p}}\)
- \(\frac{\lambda_e}{\lambda_p}=2\sqrt{\frac{m_p}{m_e}}\)
The Correct Option is D
Solution and Explanation
Electron: \(\lambda_e=\dfrac{h}{\sqrt{2m_e\,e\,V}}\); proton (same charge magnitude \(e\)): \(\lambda_p=\dfrac{h}{\sqrt{2m_p\,e\,(4V)}}\).
\(\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{2m_pe(4V)}{2m_eeV}}=\sqrt{\dfrac{4m_p}{m_e}}=2\sqrt{\dfrac{m_p}{m_e}}\).
Answer: (4) \(\dfrac{\lambda_e}{\lambda_p}=2\sqrt{\dfrac{m_p}{m_e}}\)
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