The ratio of de-Broglie wavelengths for the electron and proton moving with the same velocity is given as (m\(_e\): mass of electron, m\(_p\): mass of proton)
Show Hint
- \( m_p: m_e \)
- \( m_e: m_p \)
- \( m_e^2: m_p \)
- \( m_p^2: m_e^2 \)
The Correct Option is A
Solution and Explanation
The de-Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] Where: - \( h \) is Planck's constant, - \( p \) is the momentum, - \( m \) is the mass, - \( v \) is the velocity. For an electron and a proton moving with the same velocity \( v \): \[ \lambda_e = \frac{h}{m_e v} \] \[ \lambda_p = \frac{h}{m_p v} \] The ratio of the wavelengths is: \[ \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} \] Thus, the ratio of de-Broglie wavelengths for the electron and proton is \( m_p: m_e \).
Final Answer: \( m_p: m_e \)
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