MHT CET · Physics · Atomic Physics
The wave number of the last line of the Balmer series in hydrogen spectrum will be (Rydberg's constant \(=10^7 \mathrm{~m}^{-1}\) )
- A \(2.5 \times 10^6 \mathrm{~m}^{-1}\)
- B \(0.255 \times 10^9 \mathrm{~m}^{-1}\)
- C \(250 m^{-1}\)
- D \(2.5 \times 10^5 \mathrm{~m}^{-1}\)
Answer & Solution
Correct Answer
(A) \(2.5 \times 10^6 \mathrm{~m}^{-1}\)
Step-by-step Solution
Detailed explanation
We know that: \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
For last line Balmer's series, \(n_1=2, n_2=\propto\)
So, \(\frac{1}{\lambda}=10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=0.25 \times 10^7 \mathrm{~m}^{-1}\)
For last line Balmer's series, \(n_1=2, n_2=\propto\)
So, \(\frac{1}{\lambda}=10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=0.25 \times 10^7 \mathrm{~m}^{-1}\)
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