Question

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

• A. $0.025\times {10}^4$ $m^{-1}$
• B. $0.5\times {10}^7$ $m^{-1}$
• C. $0.25\times {10}^7$ $m^{-1}$
• D. $2.5\times {10}^7$ $m^{-1}$

Answer 1

Answer: (C)

We know that : $\frac{1}{\lambda }=R(1/n_1^2-1/n_2^2)$

For last line Balmer's series , $n_1=2,n_2=\infty$

so $\frac{1}{\lambda }={10}^7(1/2^2-1/{\infty }^2)=0.25\times {10}^7{m}^{-1}$