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) 1.32MeV
C) 59KeV
D) 13.6eV

Answer 1

Hint: The electromagnetic spectrum has some set of frequencies of radiation that is absorbed or emitted during the transitions of the electrons between the energy levels of the atoms.

Formula used:
Energy $E = \dfrac{{hc}}{\lambda }in{\text{ }}terms{\text{ }}of{\text{ }}joule$
where h is Planck’s constant, c is velocity of light, $\lambda $ is wavelength of light.

Complete step by step solution:
Consider,
Each element has a characteristic spectrum of radiation emitted by it.
Given, a line ${K_\alpha }$corresponds to the transfer of an electron from L-shell to K-shell, at $\lambda = 0.021nm$
We know that energy $E = h\nu $, ………………. (1)
We have, $c = \nu \lambda $ then $\nu = \dfrac{c}{\lambda }$
Now let us substitute the values in the equation (1)
$E = \dfrac{{hc}}{\lambda }$ in terms of Joule.
In the given option, the answer is in the form of eV. Therefore we will consider eV itself.
$E = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 0.021 \times {{10}^{ - 9}}}}$
$ = 5.9 \times {10^4}eV$

$\therefore$ The energy difference between K and L levels in this atom is about $E = 59{\text{keV}}$. Hence, the correct option is (C).

Additional information:
The electrons revolve around the nucleus of the atom in a circular orbit. The centripetal force and electrostatic force are balanced with each other then the only electron can take a circular orbit.
An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of $\dfrac{h}{{2\pi }}$ where $h$ is Planck’s constant.
That is, $mvr = \dfrac{{nh}}{{2\pi }}$
The revolving electron emits energy in the form of electromagnetic waves when it jumps from the outer stationary orbit of higher energy to the inner stationary orbit of lower energy. ${E_2} - {E_1} = h\nu $
The minimum energy required to excite an atom in the ground state to one of the higher stationary state is called excitation energy.
The minimum accelerating potential which provides all-electron energy sufficient to jump from the ground state to one of the outer orbits is called excitation potential.
The minimum energy required to ionize an atom is called the ionization energy of the atom.
The minimum accelerating potential which would provide electron energy sufficient to just remove the electron from the atom is called ionization potential.

Note:
Bohr's theory failed to explain the spectral lines that extend over the finite wavelength consisting of a small number of lines close together.