TS EAMCET · Physics · Atomic Physics
The difference in the wavelength between the maximum and minimum of Balmer series [Use \(\mathrm{R}_{\mathrm{H}}=1 \times 10^7 \mathrm{~m}^{-1}\) ]
- A \(1600 Å\)
- B \(3200 Å\)
- C \(4000 Å\)
- D \(4800 Å\)
Answer & Solution
Correct Answer
(B) \(3200 Å\)
Step-by-step Solution
Detailed explanation
For maximum wavelength of Balmer series \(\begin{aligned} & \frac{1}{\lambda_1}=10^7\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & \frac{1}{\lambda_1}=10^7\left(\frac{1}{4}-\frac{1}{9}\right) \\ & \lambda_1=\frac{36}{5} \times 10^{-7} \end{aligned}\) For minimum wavelength of…
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