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  4. Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be:

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Q. Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be:

Rajasthan PMTRajasthan PMT 1995

Solution:

de Broglie wavelength $ \lambda =\frac{12.27}{\sqrt{V}}\overset{\text{o}}{\mathop{\text{A}}}\, $ $ \lambda =\frac{12.27}{\sqrt{50000}}=0.055\overset{\text{o}}{\mathop{\text{A}}}\, $