JEE Mains · Physics · STD 12 - 12. atoms
The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly __________ nm.
- A 1875
- B 1550
- C 1217
- D 1784
Answer & Solution
Correct Answer
(C) 1217
Step-by-step Solution
Detailed explanation
N = 4 __________ N = 3 __________ paschen n = 2 __________ Balmer n = 1 __________ Lyman Smallest wavelength of lyman \( \frac{1}{\lambda}=R(\frac{1}{12}-\frac{1}{\infty^{2}}) \) \( R=\frac{1}{\lambda}=\frac{1}{91}nm^{-1} \) \( \lambda_{max} \) for balmer series…
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