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

Correct answer: A. λ_0=(2 m c λ^2)/(h) Momentum of striking electrons, p=(h)/(λ) Kinetic energy of striking electrons, K=(p^2)/(2 m)=(h^2)/(2 m λ^2) This is also, maximum energy of X -ray photons. Therefore, (h c)/(λ_0)=(h^2)/(2 m λ^2) or λ_0=(2 m λ^2 c)/(h) Correct option is (a).

JEE Advanced 2007Paper 2

Exam: JEE Advanced
Year: 2007
Paper: Paper 2
Subject: Physics
Chapter: Dual Nature of Matter

JEE Advanced 2007 · Paper 2 · Physics · Dual Nature of Matter

Electrons with de-Broglie wavelength λ fall on the target in an...

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

Options

A $\lambda_0=\frac{2 m c \lambda^2}{h}$
B $\lambda_0=\frac{2 h}{m c}$
C $\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}$
D $\lambda_0=\lambda$

Correct answer

A. $\lambda_0=\frac{2 m c \lambda^2}{h}$

Step-by-step solution

Momentum of striking electrons,

p=\frac{h}{\lambda}

\therefore

Kinetic energy of striking electrons,

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

This is also, maximum energy of

X

-ray photons. Therefore,

\quad \frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2}

\lambda_0=\frac{2 m \lambda^2 c}{h}

\therefore

Correct option is (a).

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