Physics - Modern Physics Question with Solution | TestHub

For the first transition, the electron jumps from the second excited state ($n_2=3$) to the first excited state ($n_1=2$).

Using the Rydberg formula:

$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36}$

$\lambda =\frac{36}{5R}$

For the second transition, the electron jumps from the third excited state ($n_2=4$) to the second orbit ($n_1=2$).

Let the new wavelength be ${\lambda }^{\prime }$.

$\frac{1}{{\lambda }^{\prime }}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3R}{16}$

${\lambda }^{\prime }=\frac{16}{3R}$

Taking the ratio of ${\lambda }^{\prime }$ to $\lambda$:

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{\frac{16}{3R}}{\frac{36}{5R}}=\frac{16\times 5}{3\times 36}=\frac{80}{108}=\frac{20}{27}$

${\lambda }^{\prime }=\frac{20\lambda }{27}$

Comparing this with the given expression ${\lambda }^{\prime }=\frac{20\lambda }{x}$, we get $x=27$.

Answer: $27$