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Question

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.228}

Vedclass · Accepted Answer

$9$:$4$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.228}{\sqrt{V}} 	ext{ nm}$.From this relation,we see that $\lambda \propto \frac{1}{\sqrt{V}}$,which implies $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$ or $\frac{V_1}{V_2} = \left(\frac{\lambda_2}{\lambda_1}ight)^2$.Given that the wavelength increases by $50 \%$,the new wavelength $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$.Substituting this into the ratio:$\frac{V_1}{V_2} = \left(\frac{1.5\lambda_1}{\lambda_1}ight)^2 = \left(\frac{3}{2}ight)^2 = \frac{9}{4}$.

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is

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