Electrons of mass m with de-Broglie wavelength λ fall on the target. The cut-off wavelength λ_0 is equal to [ h= Planck's constant, C= velocity of light ]

Correct answer: (A) (2 m c λ^2)/h. Using de-Broglie equation λ=h/p where p=√(2 m E) ⇒ λ=h/(√(2 m E)) Energy of the X-ray emitted E=(h c)/(λ_0) ∴ λ=h/(√(2 m × (h c)/(λ_0))) ⇒ λ_0=(2 m c λ^2)/h

Q114 / 150 · MHT CET 2022 (08 Aug Shift 1) · Physics

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target. The cut-off wavelength $\lambda_0$ is equal to [ $h=$ Planck's constant, $C=$ velocity of light $]$

Correct answer
A
$\frac{2 m c \lambda^2}{h}$
B
$\frac{m c \lambda}{h}$
C
$\frac{2 h}{m c \lambda^2}$
D
$\frac{2 m c \lambda}{h}$

Solution

Using de-Broglie equation $\lambda=\frac{h}{p}$ where $p=\sqrt{2 m E}$ $\Rightarrow \lambda=\frac{h}{\sqrt{2 m E}}$ Energy of the X-ray emitted $E=\frac{h c}{\lambda_0}$ $\therefore \lambda=\frac{h}{\sqrt{2 m \times \frac{h c}{\lambda_0}}} \Rightarrow \lambda_0=\frac{2 m c \lambda^2}{h}$

Hence, the correct answer is option: 'A'.

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