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

(A) $\lambda_{0}=\frac{2 m c \lambda^{2}}{h}$

(B) $\lambda_{0}=\frac{2 h}{m c}$

(C) $\lambda_{0}=\frac{2 m^{2} c^{2} \lambda^{3}}{h^{2}}$

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Modern Physics

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

(A) $λ_{0} = \frac{2 m c λ ^{2}}{h}$ ✓ Correct
(B) $λ_{0} = \frac{2 h}{m c}$
(C) $λ_{0} = \frac{2 m ^{2} c ^{2} λ ^{3}}{h ^{2}}$
(D) $λ_{0} = λ$

Answer: (A)

Solution

The de-Broglie wavelength of electrons is

λ = \frac{h}{p} = \frac{h}{2 m ( e V )}

Solving for the accelerating potential:

e V = \frac{h ^{2}}{2 m λ ^{2}}

The cut-off wavelength of X-rays corresponds to maximum photon energy, which equals the kinetic energy of electrons:

E_{ma x} = e V = \frac{h c}{λ _{0}}

Therefore:

\frac{h c}{λ _{0}} = \frac{h ^{2}}{2 m λ ^{2}}

λ_{0} = \frac{h c \cdot 2 m λ ^{2}}{h ^{2}} = \frac{2 m c λ ^{2}}{h}

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option

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