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12. atoms — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics12. atoms1 Mark
MCQ
An electron having de-Broglie wavelength $\lambda$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted $X$-ray is :
  • A
    ${\lambda _0} = \frac{{2{m^2}{c^2}{\lambda ^3}}}{{{h^2}}}$
  • B
    $\;{\lambda _0}$ $= λ$
  • ✓
    $\;{\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}$
  • D
    $\;{\lambda _0} = \frac{{2h}}{{mc}}$
✓

Answer

Correct option: C.
$\;{\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}$
c
The correct option is $D \frac{2 m c \lambda^2}{h}$ Let $K$ be the kinetic energy of the electron.

Electron de- Broglie wavelength

$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m K}}$

$\Rightarrow K=\frac{h^2}{2 m \lambda^2}$

Now, cut- off wavelenth for continuous $X$ - rays,

$\lambda_c=\frac{h c}{K}=\frac{h c}{\left(h^2 / 2 m \lambda^2\right)}$

$\therefore \lambda_c=\frac{2 m c \lambda^2}{h}$

Hence, option $(C)$ is correct.

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