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MHT CET · Physics · Atomic Physics

The wave number of the last line of the Balmer series in hydrogen spectrum will be (Rydberg's constant \(=10^7 \mathrm{~m}^{-1}\) )

  1. A \(2.5 \times 10^6 \mathrm{~m}^{-1}\)
  2. B \(0.255 \times 10^9 \mathrm{~m}^{-1}\)
  3. C \(250 m^{-1}\)
  4. D \(2.5 \times 10^5 \mathrm{~m}^{-1}\)
Verified Solution

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, \(\frac{1}{\lambda}=10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=0.25 \times 10^7 \mathrm{~m}^{-1}\)