SATHEE: Modern Physics 4 Question 12

Modern Physics 4 Question 12

12. Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted $X$ -rays is

(2007, 3M)

(a) $λ_{0} = \frac{2 m c λ^{2}}{h}$

(b) $λ_{0} = \frac{2 h}{m c}$

(d) $λ_{0} = λ$

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Solution:

Momentum of striking electrons

$p = \frac{h}{λ}$

$∴$ Kinetic energy of striking electrons

$K = \frac{p^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$

This is also, maximum energy of $X$ -ray photons.

Therefore, $\frac{h c}{λ_{0}} = \frac{h^{2}}{2 m λ^{2}}$ or $λ_{0} = \frac{2 m λ^{2} c}{h}$

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Modern Physics 4 Question 12

12. Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

Solution:

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12. Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted $X$ -rays is