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  4. An electron accelerated through a potential difference V1 has a de-Broglie wavelength of λ. When the potential is changed to V2, its de-Broglie wavelength increases by 50 %. The value of ((V1/V2)) is equal to

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Q. 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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Solution:

$ KE =\frac{ P ^2}{2 m }, $
$P =\frac{ h }{\lambda} $
$ eV _1=\frac{\left(\frac{ h }{\lambda}\right)^2}{2 m } $
$ eV _2=\frac{\left(\frac{ h }{1.5 \lambda}\right)^2}{2 m } $
$ \frac{ V _1}{ V _2}=(1.5)^2=\frac{9}{4}$