Question

Which of the following transitions results in the maximum frequency of emission?
A. n = 2 to n = 1
B. n = 6 to n = 2
C. n = 1 to n = 2
D. n = 2 to n = 6

Answer 1

Hint: This question is related to Bohr’s model of a Hydrogen Atom, in which the frequency of transition from one orbit ($n_1$) to another ($n_2$) can be derived by:
$\begin{align}
  & \dfrac{1}{\lambda }=(\dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}}){{R}_{H}},\text{ where }{{\text{R}}_{\text{H}}}\text{ is the Rydberg Constant = 1}\text{.09678 X 1}{{\text{0}}^{\text{-2}}}\text{n}{{\text{m}}^{\text{-1}}}\text{.} \\
 & \Rightarrow \dfrac{\upsilon }{c}=(\dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}}){{R}_{H}}(\because \lambda =\dfrac{c}{\upsilon })\text{ , where c is the speed of light}\text{.} \\
 & \Rightarrow \upsilon \text{=}(\dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}}){{R}_{H}}*c \\
\end{align}$
Where $\lambda $ is the wavelength of transmission and $\upsilon $ is its frequency.

Complete step-by-step solution:
Let us now observe the expression derived in the hint.
Since all other terms except the energy levels of the two orbits are constant, we can determine that the maximum frequency transmission will be in that scenario in which the electron makes the largest transition from a lower energy level to a higher energy level as a transition from a higher energy level to a lower energy level would result in negative frequency.
Thus, we eliminate options a) and b).
Of the other two possible answers, the transition from n = 1 to n = 2 is a larger transition than when compared to n = 2 to n = 6 according to the above formula since for the transition from n = 1 to n = 2:
$\Rightarrow \upsilon \text{=}0.75{{R}_{H}}*c$
However, for the transition from n= 2 to n = 6:
$\Rightarrow \upsilon \text{=}0.22{{R}_{H}}*c$
Therefore, we conclude that the transition from n = 1 to n = 2 is the greatest from the given options, making the answer to this question C).

NOTE: It is not required to calculate the exact frequency of each of the given options to be able to solve this question, as simple observation of the wavelength equation will do the same trick.
Please do not be fooled by a larger numerical transition as shown in option d) and make sure you have analysed each and every option according to the required formula before giving an answer.