Question

In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :

• A. $\frac{27}{20} \lambda $
• B. $\frac{16}{25} \lambda $
• C. $\frac{20}{27} \lambda $
• D. $\frac{25}{16} \lambda $

Hint:

You do not need to know the value of K - set up both transitions as one ratio and it cancels out on its own.

Answer 1

The Correct Option is C

Concept:

• Since the same proportionality constant appears in both transitions, set the two Rydberg expressions up as a single ratio from the start instead of solving for each wavelength separately and then dividing.

Step 1: Write both transitions using the Rydberg formula, keeping the constant $K$ symbolic.
$M\to L$ (shell 3 to shell 2): $\dfrac{1}{\lambda} = K\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)$
$N\to L$ (shell 4 to shell 2): $\dfrac{1}{\lambda_N} = K\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)$

Step 2: Divide the two equations directly so $K$ cancels immediately.
$\dfrac{1/\lambda_N}{1/\lambda} = \dfrac{\lambda}{\lambda_N} = \dfrac{\frac14-\frac1{16}}{\frac14-\frac19}$

Step 3: Simplify each bracket.
$\dfrac14-\dfrac{1}{16} = \dfrac{3}{16}$, and $\dfrac14-\dfrac19 = \dfrac{5}{36}$
$\dfrac{\lambda}{\lambda_N} = \dfrac{3/16}{5/36} = \dfrac{3}{16}\times\dfrac{36}{5} = \dfrac{108}{80} = \dfrac{27}{20}$

Step 4: Flip to get $\lambda_N$ in terms of $\lambda$.
$\lambda_N = \dfrac{20}{27}\lambda$

Final Answer: $\lambda_N = \dfrac{20}{27}\lambda$