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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Given the ratio of the de-Broglie wavelength of the particle $(p)$ to

Vedclass · Accepted Answer

$\frac{1}{8}$ times the mass of $e^{-}$

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Given the ratio of the de-Broglie wavelength of the particle $(p)$ to the electron $(e)$ is $\frac{\lambda_p}{\lambda_e} = 2:1$.Also,the velocity of the particle is $v_p = 4v_e$.Using the formula $\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p}$,we substitute the known values:$2 = \frac{m_e v_e}{m_p (4v_e)}$Simplifying the equation:$2 = \frac{m_e}{4m_p}$Solving for $m_p$:$m_p = \frac{m_e}{4 	imes 2} = \frac{m_e}{8}$.Therefore,the mass of the particle is $\frac{1}{8}$ times the mass of the electron.

$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

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