Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of 0.

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Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of 0. If the voltage is increased to 40 kV, then the de-Broglie wavelength associated…

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Given: An electron beam has de-Broglie wavelength $\lambda_0$ at accelerating voltage $20 \, \text{kV}$ .

Find: The new de-Broglie wavelength when the accelerating voltage becomes $40 \, \text{kV}$ .

For an electron accelerated through a potential difference $V$ , the de-Broglie wavelength is

\lambda = \frac{h}{\sqrt{2meV}}

where $h$ is Planck's constant, $m$ is the mass of the electron, and $e$ is the electronic charge.

Hence,

\lambda \propto \frac{1}{\sqrt{V}}

If at $V_1 = 20 \, \text{kV}$ , the wavelength is $\lambda_1 = \lambda_0$ , then at $V_2 = 40 \, \text{kV}$ ,

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{20}{40}} = \frac{1}{\sqrt{2}}

So,

\lambda_2 = \lambda_0 \cdot \frac{1}{\sqrt{2}} = \frac{\lambda_0}{\sqrt{2}}

Therefore, the de-Broglie wavelength becomes $\frac{\lambda_0}{\sqrt{2}}$ . The correct option is D.

Note: The solution marks option C, but its working concludes $\frac{\lambda_0}{\sqrt{2}}$ , which matches option D in the given options.

Given: The voltage is doubled from 20 kV to 40 kV. Find: How the de-Broglie wavelength changes. Since 1 V doubling V makes the wavelength smaller by a factor of 2. So the new wavelength is = 0 2 Therefore, the correct option is D.

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