Question

If $\Delta E$ is the energy emitted in electron volts when an electronic transition occurs from higher energy level to a lower energy level in H-atom, the wavelength of the line produced is approximately equal to
A. $\dfrac{19800}{\Delta E}\overset{\text{o}}{\mathop{\text{A}}}\,$
B. $\dfrac{12375}{\Delta E}\overset{\text{o}}{\mathop{\text{A}}}\,$
C. $\dfrac{13600}{\Delta E}\overset{\text{o}}{\mathop{\text{A}}}\,$
D. $\dfrac{21800}{\Delta E}\overset{\text{o}}{\mathop{\text{A}}}\,$

Answer 1

Hint: There is a relation between $\Delta E$ and the wavelength and it is as follows.
\[\Delta E=hv=\dfrac{hc}{\lambda }\]
Here $\Delta E$ = change in energy
h = Planck’s constant
c = velocity of the light
$\lambda $ = wavelength of the light

Complete answer:
- In the question it is asked to find the wavelength of the line which is going to be produced when an electron is going to jump from higher energy level to lower energy level.
- The electron which is going to jump from higher energy level to lower energy level belongs to hydrogen atom as per the question.
- There is a formula to calculate the wavelength of the line and it is as follows.
\[\begin{align}
  & \Delta E=hv=\dfrac{hc}{\lambda } \\
 & \lambda =\dfrac{hc}{\Delta E} \\
\end{align}\]
Here $\Delta E$ = change in energy
h = Planck’s constant = $6.6\times {{10}^{-34}}kg{{m}^{2}}/s$
c = velocity of the light = $3\times {{10}^{8}}m/s$
$\lambda $ = wavelength of the light
- Substitute all the known values in the formula to get the wavelength of the line and it is as follows.
\[\begin{align}
  & \Delta E=hv=\dfrac{hc}{\lambda } \\
 & \lambda =\dfrac{hc}{\Delta E} \\
 & \lambda =\dfrac{(6.6\times {{10}^{-34}}kg{{m}^{2}}/s)(3\times {{10}^{8}}m/s)}{\Delta E} \\
 & \lambda =\dfrac{19.8\times {{10}^{-26}}}{\Delta E} \\
\end{align}\]
- In the above calculation $\Delta E$is in joules, we have to convert it into eV and it is as follows.
- We know that 1 eV = $1.6\times {{10}^{-19}}J$ .
\[\begin{align}
  & \lambda =\dfrac{19.8\times {{10}^{-26}}}{\Delta E} \\
 & \lambda =\dfrac{19.8\times {{10}^{-26}}}{\Delta E\times 1.6\times {{10}^{-19}}} \\
 & \lambda =\dfrac{12375\times {{10}^{-10}}}{\Delta E}m \\
 & \lambda =\dfrac{12375}{\Delta E}\overset{\text{o}}{\mathop{\text{A}}}\, \\
\end{align}\]

Therefore, the correct option is B.

Note:
We have to convert the energy value in electron volts (eV) for that we have to use the relationship between electron volt and joules. The relation between the electron volt (eV) and joules is as follows.
1 eV = $1.6\times {{10}^{-19}}J$ .