- Home
- Standard 12
- Physics
12.Atoms
normal
Electrons with de-Broglie wavelength $\lambda $ fall on the target an $X-$ ray tube. The cut off wavelength of emitted $X-$ ray is
A
${\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}$
B
${\lambda _0} = \frac{{2h}}{{mc}}$
C
${\lambda _0} = \frac{{2{m^2}{c^2}{\lambda ^3}}}{{{h^2}}}$
D
${\lambda _0} = \lambda $
Solution
for striking $\mathrm{e}^{-}: \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \therefore \mathrm{eV}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda^{2}}$
for $\mathrm{X}$ -ray $: \lambda_{\min }=\frac{\mathrm{hc}}{\mathrm{eV}}=\frac{\mathrm{hc}}{\mathrm{h}^{2} / 2 \mathrm{m} \lambda^{2}}=\frac{2 \mathrm{mc} \lambda^{2}}{\mathrm{h}}$
Standard 12
Physics
Similar Questions
normal
normal
