Question

The $K_{\alpha }$ X-ray emission line of tungsten occurs at $\lambda =0.021nm$. The energy difference between K and L levels in this atom is about

• A. $0.51MeV$
• B. $1.2MeV$
• C. $59keV$
• D. $13.6eV$

Answer 1

Answer: (C)

$59keV$

Line $K_{\alpha }$ corresponds to transfer of electron from L-shell to K-shell.
Here, $\lambda =0.021nm=2.1\times {10}^{-11}m$

We know that $\lambda =\frac{hc}{E\left(inJoule\right)}$
$\lambda =\frac{hc}{E\left(inJoule\right)}=\frac{hc}{e\times E\left(ineV\right)}$
$\Rightarrow E\left(ineV\right)=\frac{hc}{e\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 2.1\times {10}^{-11}}eV$
$E\left(ineV\right)=5.9\times {10}^{4}eV=59keV$