(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sq

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Wavelength is decreased to $\frac{1}{\sqrt{2}}$ times.

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Let the initial potential difference be $V_1 = V$ and the final potential difference be $V_2 = 2V$.The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.Therefore,the de-Broglie wavelength decreases by a factor of $\frac{1}{\sqrt{2}}$.

Class 12 Physics Dual Nature of Radiation and matter Matter Waves and de Broglie Wavelength

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If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

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A
Wavelength is increased to $\sqrt{2}$ times.
B
Wavelength is decreased to $\sqrt{2}$ times.
C
Wavelength is decreased to $\frac{1}{\sqrt{2}}$ times.
D
Wavelength is increased to $\frac{1}{\sqrt{2}}$ times.

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Matter Waves and de Broglie Wavelength

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If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

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The kinetic energy of an electron is increased by $2$ times,then the de-Broglie wavelength associated with it changes by a factor.

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The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$ is $n:1$. The value of $n$ is . . . . . . .
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