Question

A potential difference of $20\mathrm{kV}$ is applied across an X-ray tube. The minimum wavelength of X-rays generated is ____________ ${\mathrm{A}}^{\circ }$.

• A. $1.621\mathrm{A}°$
• B. $0.621\mathrm{A}°$
• C. $2.621\mathrm{A}°$
• D. $5.621\mathrm{A}°$

Answer 1

The correct option is B

$0.621\mathrm{A}°$

Explanation for correct option

Step 1: Given data

Potential difference $\mathrm{V}=20\mathrm{kV}$

$=20000\mathrm{V}$

$\mathrm{e}=1.6\times {10}^{-19}\mathrm{Columbus}$

$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

Step 2: Find the wavelength

The amount of energy obtained by a charged particle when it is accelerated through a potential $\mathrm{V}$ is,

$\mathrm{E}=\mathrm{eV}$ ----------- $\left(1\right)$

Here, $\mathrm{e}$ is charge, and $\mathrm{V}$ is potential.

The wavelength of the emitted X-ray is,

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{eV}}$ ---------- $\left(2\right)$

Here, $\mathrm{h}$ is planks constant, $\mathrm{\lambda }$ is wavelength, $\mathrm{e}$ is charge and $\mathrm{c}$ is velocity of light.

Substitute the given values in equation $\left(2\right)$. We get,

$\mathrm{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 20000}$

$=\frac{1.98\times {10}^{-25}}{3.2\times {10}^{-15}}$

$=0.62\times {10}^{-10}\mathrm{m}$

$=0.62\mathrm{A}°$

Therefore, the wavelength of x-ray generated is $0.62\mathrm{A}°$

Hence, option $\mathrm{B}$ is correction answer.