Question

Electronic transition in $H{{e}^{+}}$ would emit light in which region when ${{e}^{-}}$ jumps from fourth orbit to second orbit:
A. UV
B. Visible
C. IR
D. No spectral region can be defined

Answer 1

Hint: Molecular electronic transitions take place when electrons in a molecule are excited from one energy level to a higher energy level. The energy change during this transition gives us information about the structure of a molecule and also tell about some other molecular properties such as color.

Complete step by step solution:
The relationship between the energy involved in the electronic transition and the frequency of radiation is given by Planck's relation. Planck’s relation is shown by:
$E=\dfrac{h}{nu} $ and it can also be expressed as:
$E=h\dfrac{c}{\lambda }$
Energy radiated when $H{{e}^{+}}$ jumps from 4th to 2nd orbit is given by formula:
$E=-13.6{{Z}^{2}}[\dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}}]$
For He, Z = 2, ${{n}_{1}}$= 2 and ${{n}_{2}}$= 4
Put the values in the formula
$E=-13.6{{(2)}^{2}}[\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}}]$
$E=-13.6\times 4[\dfrac{1}{4}-\dfrac{1}{16}]$
$E=-13.6\times 4[\dfrac{4-1}{16}]$
$E=-13.6\times 4\times \dfrac{3}{16}$
$\therefore -10.2 ev/atom$
Radiation emit in which orbital it can be calculated with the help of planck's constant
i.e. $E=h\dfrac{c}{\lambda }$
$\lambda =\dfrac{1240}{E(ev)}nm$
$\lambda =\dfrac{1240}{10.2}nm$
$\lambda =121.568nm$
Hence the wavelength is given as 121.568 nm and we know that this wavelength lies in the region of UV i.e. ultraviolet region.

Thus, option A is the correct answer.

Note: UV stands for ultraviolet region and it can be explained as a form of electromagnetic radiation having wavelength from 10 to 400 nm and the wavelength given by UV is shorter than that of visible light but it is longer than X-rays.