Modern Physics Ques 152
- 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
(2007, 3M)
(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) $\lambda _0=\lambda$
Show Answer
Answer:
Correct Answer: 152.(a)
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, $\frac{h c}{\lambda _0}=\frac{h^{2}}{2 m \lambda^{2}} \quad$ or $\quad \lambda _0=\frac{2 m \lambda^{2} c}{h}$
BETA
