The wavelength of the radiation emitted is \(λ_0\) when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be \((20λ_0/x)\). The value of \(x\) is
Show Hint
Use \(\frac1\lambda=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\); second excited state is \(n=3\) and third excited is \(n=4\).
Step 1: Plan:
Take the ratio of the two wave numbers directly.
Step 2: Steps:
$\frac{1/\lambda}{1/\lambda_0} = \frac{3R/16}{5R/36} = \frac{3}{16}\times\frac{36}{5} = \frac{27}{20}$. So $\frac{\lambda}{\lambda_0} = \frac{20}{27}$ and $x = 27$.
A common slip is taking the third excited state as $n=3$, which would make the two lines identical.
Final Answer:
The value of $x$ is $27$, option (C).
\[ \boxed{27} \]