Electrons of mass m with de-Broglie wavelength \lambda fall on the target in an X-ray tube. The cutoff wavelength (\lambda_0) of the emitted X-ray is: [NEET 2016 P2]

Correct Answer is : λ0=2mcλ2h

\lambda_0={2m^2c^2\lambda^3}/h^2

NEET 2016 Physics and Chemistry Paper 2

Question : 48 of 90

Marks: +1, -0

Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ0) of the emitted X-ray is:

[NEET 2016 P2]

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