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

(B) $\frac{2 m c \lambda^{2}}{h}$

(C) $\frac{\mathrm{hc}}{\mathrm{mc}}$

(D) $\frac{2 m^{2} c^{2} \lambda^{2}}{h^{2}}$

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Modern Physics

An electron having de-Broglie wavelength

λ

is indent on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:

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

Answer: (B)

Solution

The incident electron carries momentum

p = \frac{h}{λ}

, so its kinetic energy is

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

At the cut-off the whole of this energy goes into a single photon:

\frac{h c}{λ _{m i n}} = \frac{h ^{2}}{2 m λ ^{2}}

The cut-off wavelength of the emitted X-ray is therefore

\frac{2 m c λ ^{2}}{h}

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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