Question

The $K_{\alpha }$ x-ray emission line of tungsten occurs at $\lambda$ = 21 pm. What is the energy difference between $E_K$ and $E_L$ levels in this atom (MeV refers to mega electron volts;1MeV = ${10}^6$ eV and hc = 1242 eVnm)?

• A. 0.59 MeV
• B. 1.2 MeV
• C. 59 KeV
• D. 13.6 eV

Answer 1

The correct option is A 0.59 MeV

$K_{\alpha }$ x-ray emission takes place when an electron jumps from L-shell to fill a vacancy created in K-shell.
Now $E_K$ is the energy of the atom when vacancy is in K-Shell, and $E_L$ Is the energy of the atom when vacancy is created in L-Shell. Also, $E_K$ > $E_L$ So, when vacancy shifts from K to L, according to our agrument, the atomic energy decreases. How much ?
$E_K-E_L$.
This difference results into X-ray photon.
Hence connecting the dots we can conclude that ; The energy difference between K and L levels is the energy of the photon.
Mathematically;
$E_K-E_L=E_{photon}$

Now,
$E_K-E_L=\frac{hc}{\lambda }$

Which gives after putting in the values;
$E_K-E_L=\frac{1242eVnm}{0.0021nm}$

Final calculations leads us to our answer;
$E_K-E_L=591428eV$
Or,
$E_K-E_L=0.59MeV$