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

We need to understand the relation between wavelength of the emitted spectrum of an element and the energy difference involved in the two levels of the shells in an electron system. We can use this relation to find the energy difference.

Complete answer:
We are given the wavelength of X-ray emitted from the tungsten element. We know that the emission of an electromagnetic wave occurs when the electron in an excited state in a higher energy level moves down to a lower energy level. The de-excitation of the electron from the higher state to the lower state is accompanied by the release of an electromagnetic wave which has the energy equivalent to the energy difference between the two levels involved in this.
Here, we are given the de-excitation of an electron in a tungsten atom from the L-level to the K-level releasing an X-ray of wavelength $\lambda =0.021nm$. The energy difference between the two levels can be easily calculated by using the formula which relates the energy of a wave and its frequency. It is given as –
$E=h\nu$
Where, E is the energy difference between the levels K and L,
h is the planck's constant with the value $h=6.626\times {{10}^{-34}}Js$
$\nu$ is the frequency of the emitted radiation.
We know that the frequency of a radiation is related to wavelength by the formula –

& \upsilon =\dfrac{c}{\lambda } \\\ & \therefore E=\dfrac{hc}{\lambda } \\\ \end{aligned}$$  We can find the energy by substituting the values for the given parameters and convert it into the units of electron volts as – $$\begin{aligned} & E=\dfrac{hc}{\lambda } \\\ & \Rightarrow E=\dfrac{6.626\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m{{s}^{-1}}}{2.1\times {{10}^{-11}}m} \\\ & \therefore E=9.47\times {{10}^{-15}}J \\\ & E(eV)=\dfrac{E(J)}{e} \\\ & \Rightarrow E=\dfrac{9.47\times {{10}^{-15}}J}{1.6\times {{10}^{-19}}C} \\\ & \therefore E=59187.2eV=59keV \\\ \end{aligned}$$ This is the energy difference between the K and L levels in a tungsten atom. **The correct answer is option C.** **Note:** The wavelength of the electromagnetic radiation released by the de-excitation of an electron in an atom is always a discrete value. The energy is released as quanta or photons which will carry the details about the atom without any loss of data.