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(C) Let the two particles have charges $q_1$ and $q_2$. Given $q_1 = q_2 = q$. Both are accelerated through the same potential difference $V$.The kine

Vedclass · Accepted Answer

$\lambda_1 : \lambda_2 = \sqrt{m_2} : \sqrt{m_1}$

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(C) Let the two particles have charges $q_1$ and $q_2$. Given $q_1 = q_2 = q$. Both are accelerated through the same potential difference $V$.The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through potential $V$ is $K = qV$.Since $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK} = \sqrt{2mqV}$.The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.Since $h$,$q$,and $V$ are constant for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.Therefore,the ratio of their wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}}$.

Two particles have the same charge. If they are accelerated through the same potential difference,what will be the ratio of their de Broglie wavelengths?

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