Physics · Atomic Physics
JEE Main 2024 — 27 January, Shift 2 — Question 58
If Rydberg's constant is , the longest wavelength of radiation in Paschen series will be , where .
Answer: 144
Numerical answer — enter this value.
Step-by-step solution
Longest wavelength corresponds to transition between and
$\begin{array}{*{20}{r}}{}&{\frac{1}{\lambda } = {\rm{R}}{{\rm{Z}}^2}\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) = {\rm{R}}{{\rm{Z}}^2}\left( {\frac{1}{9} - \frac{1}{{16}}} \right)}\\{}&{\; = \frac{{7R{Z^2}}}{{9 \times 16}}}\\{}&{\; \Rightarrow \lambda = \frac{{144}}{{7R}}{\rm{\;for\;}}Z = 1{\rm{\;}}\therefore \alpha = 144}\end{array}$
Answer key and solution verified before publishing.
Practise Atomic Physics
Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.
- Exam
- JEE Main 2024
- Paper
- 27 January, Shift 2
- Subject
- Physics
- Chapter
- Atomic Physics
- Topic
- Hydrogen Spectrum