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(D) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.S

Vedclass · Accepted Answer

$43: 1$

Vedclass · Answer

(D) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Since $h$,$q$,and $V$ are the same for both particles,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.Therefore,the ratio of the wavelengths is $\frac{\lambda_{e}}{\lambda_{P}} = \sqrt{\frac{m_{P}}{m_{e}}}$.Substituting the given values: $\frac{\lambda_{e}}{\lambda_{P}} = \sqrt{\frac{1.00727}{0.00055}} \approx \sqrt{1831.4} \approx 42.79$.Rounding this value,we get approximately $43: 1$.

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through the same potential of $100\, V$. What should nearly be the ratio of their wavelengths? $(m_{P} = 1.00727\, u, m_{e} = 0.00055\, u)$

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