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JEE · NEET Previous Year Question

$An electron having de-Broglie wavelength λ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:$

Topic: JEE PYQ

Answer: C. Full step-by-step solution below.

A $0$
B $\frac{2 m ^{2} c ^{2} λ ^{2}}{h ^{2}}$
C $\frac{2 mc λ ^{2}}{h}$
D $\frac{hc}{mc}$

Correct answer: C

Solution

(c) λ = \frac{h}{mV} kinetic energy, \frac{P ^{2}}{2 m} = \frac{h ^{2}}{2 m λ ^{2}} = \frac{hc}{λ _{c}} λ_{C} = \frac{2 m λ ^{2} c}{h}

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