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An electron microscope is designed to study very small objects like viruses, microbes, and the crystal structure of solids. What is the De-Broglie wavelength of the electrons utilized by the microscope, given that they are accelerated by a potential difference of 500 V?

Option: 1

0.082 \AA


Option: 2

0.123 \AA


Option: 3

0.246 \AA


Option: 4

0.492 \AA


Answers (1)

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The De-Broglie wavelength of an electron can be calculated using the formula \lambda_{e}= \frac{h}{\sqrt{2m\left ( eV \right )}}= \frac{12.27}{\sqrt{V}} \AA, where h is Planck's constant,  m is the mass of the electron, The symbol e represents the charge of the electron, and V represents the potential difference that the electron is accelerated through.

Substituting the given value of V=500 V, we get \lambda_{e}= \frac{12.27}{\sqrt{500}} \AA \approx 0.123\, \AA.

Therefore, the correct option is (2).

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avinash.dongre

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