Question

Given the value of Rydberg Constant is ${10}^{-7}{m}^{-1}$, the wave number of last line of the Balmer series in hydrogen spectrum will be:

• A. $0.025\times {10}^4{m}^{-1}$
• B. $0.25\times {10}^{-7}{m}^{-1}$
• C. $2.5\times {10}^{-7}{m}^{-1}$
• D. $2.5\times {10}^7{m}^{-1}$

Answer 1

Answer: (B), (D)

• The Rydberg equation is given as:$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\overline{v}=wavenumber$
• For the last line in the Balmer's series​$n_1=2and{n}_2=\infty$
• The value of the Rydberg constant is given $R={10}^7$
• Hence, the wavenumber of the last line of the Balmer series can be calculated as:

  $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\overline{v}$

  $\frac{1}{\lambda }={10}^7(\frac{1}{2^2}-\frac{1}{{\infty }^2})$

  $=0.25\times 10m^{-1}$