Physics - Modern Physics Question with Solution | TestHub

Key Idea Number of spectral lines abtained due to transition of electron from ${\mathbf{n}}^{\text{th }}$ orbit to lower orbit is $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum. Number of spectral lines $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ $$\begin{array}{rlr}\Rightarrow & \frac{n(n-1)}{2}& =6\\ \text{ or }& n^2-n-12& =0\\ \text{ or }& (n-4)(n+3)& =0\\ \text{ or }& n& =4\end{array}$$ Now as the first line of the series has the maximum wavelength, therefore electron jumps from the $4^{\text{th }}$ orbit to the third orbit.