Exam

JEE Main

Year

2024

Date

5 April

Shift

Evening Shift

Subject

Physics

Chapter

Atomic Physics

JEE Main 2024 · 5 April · Evening Shift · Physics · Atomic Physics

The shortest wavelength of the spectral lines in the Lyman series...

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $915 Ã…$. The longest wavelength of spectral lines in the Balmer series will be _____ $Ã…$.

Step-by-step solution

Shortest, $\frac{\mathrm{hc}}{\lambda}=-13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \lambda \downarrow \mathrm{E} \uparrow ; \frac{\mathrm{hc}}{\lambda_0}=-13.6(1)$ Balmer Series :
$$\begin{aligned} & \frac{\mathrm{hc}}{\lambda_1}=-13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & \frac{\mathrm{hc}}{\lambda_1}=-13.6\left(\frac{1}{4}-\frac{1}{9}\right) \end{aligned}$$$$\begin{aligned} & \frac{\mathrm{hc}}{\lambda_1}=-13.6 \times\left(\frac{5}{36}\right) \\ & \Rightarrow \frac{-13.6 \lambda_0}{\lambda_1}=-13.6 \times \frac{5}{36} \\ & \lambda_1=\frac{\lambda_0 \times 36}{5}=\frac{915 \times 36}{5}=6588\end{aligned}$$$