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(A) The initial kinetic energy of the electron is $K_i = 100 \, eV$.When accelerated through a potential difference of $V = 50 \, V$,the electron gain

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$1$

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(A) The initial kinetic energy of the electron is $K_i = 100 \, eV$.When accelerated through a potential difference of $V = 50 \, V$,the electron gains additional kinetic energy equal to $K' = eV = 50 \, eV$.The total kinetic energy of the electron becomes $K_f = K_i + K' = 100 \, eV + 50 \, eV = 150 \, eV$.The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$.Using the shortcut formula $\lambda (	ext{in } \mathring{A}) = \sqrt{\frac{150}{K (	ext{in } eV)}}$,we get:$\lambda = \sqrt{\frac{150}{150}} = \sqrt{1} = 1 \, \mathring{A}$.

An electron with an initial kinetic energy of $100 \, eV$ is accelerated through a potential difference of $50 \, V$. The final de-Broglie wavelength of the electron becomes .................... $\mathring{A}$

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