Question

What should be the approximate kinetic energy of an electron so that its de-Broglie wavelength is equal to the wavelength of X-ray of maximum energy, produced in an X-ray tube operating at $24800\text{V}$?
(Given that $h=6.6\times$ 10${}^{-34}$ Joule-sec, mass of electron = 9.1$\times$${10}^{-31}\text{kg}$)

• A. $600\text{eV}$
• B. $365\text{eV}$
• C. $120\text{eV}$
• D. $300\text{eV}$

Answer 1

The correct option is A $600\text{eV}$

Mass of electron, $m=9.1\times {10}^{-31}$ kg.
Energy produced = 24800 eV
$h=6.6\times {10}^{-34}$ J-sec.

We know that $E=h\nu =\frac{hc}{\lambda }$
$\Rightarrow \lambda =\frac{hc}{E}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{24800\times 1.6\times {10}^{-16}}$
$\Rightarrow \lambda =0.5\times {10}^{-10}\text{m}$
Now from de-Broglie wavelength formula,
$\lambda =\frac{h}{P}\Rightarrow P=\frac{h}{\lambda };$
where $P$ is momentum
$\frac{6.6\times {10}^{-34}}{0.5\times {10}^{-10}}=13.2\times {10}^{-24}$
Now energy, $E=\frac{P^2}{2m}=\frac{(13.2\times {10}^{-24}{)}^2}{2\times 9.1\times {10}^{-31}}$
$E=6\times {10}^{2}eV=600eV$