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

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$\frac{9}{4}$

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(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.This implies $\lambda \propto \frac{1}{\sqrt{V}}$,or $V \propto \frac{1}{\lambda^2}$.For the first case,$V_1 \propto \frac{1}{\lambda^2}$.In the second case,the wavelength increases by $50 \%$,so the new wavelength $\lambda' = \lambda + 0.5\lambda = 1.5\lambda = \frac{3}{2}\lambda$.Thus,$V_2 \propto \frac{1}{(\frac{3}{2}\lambda)^2} = \frac{1}{\frac{9}{4}\lambda^2}$.Taking the ratio,$\frac{V_1}{V_2} = \frac{1/\lambda^2}{1/(\frac{9}{4}\lambda^2)} = \frac{9}{4}$.

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. 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 equal to :

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