Question

The first emission line of hydrogen atomic spectrum in the Balmer series appears at (R = Rydberg constant):

• A. $\frac{5R}{36}c{m}^{-1}$
• B. $\frac{3R}{4}c{m}^{-1}$
• C. $\frac{7R}{144}c{m}^{-1}$
• D. $\frac{9R}{400}c{m}^{-1}$

Answer 1

The correct option is A $\frac{5R}{36}c{m}^{-1}$

Use the formula ,
wavenumber =R×(1/n21−1/n22)=R×(1/n12−1/n22).