Question

What is maximum wavelength for Lymen series and Balmer series for H-spectrum respectively.

(1) \(\left( \frac{4}{3R_H} \right) , \left( \frac{36}{5R_H} \right)\)

(2) \(\left( \frac{36}{5R_H} \right) , \left( \frac{4}{3R_H} \right)\)

(3) \(\left( \frac{36}{7R_H} \right) , \left( \frac{3}{4R_H} \right)\)

(4) \(\left( \frac{3}{4R_H} \right) , \left( \frac{36}{7R_H} \right)\)

Answer 1

Correct option is:  (1) \(\left( \frac{4}{3R_H} \right) , \left( \frac{36}{5R_H} \right)\)

\(\frac{1}{\lambda} = R_H \left[\frac{1}{n_1 ^2 - \frac{1}{n_2 ^2}} \right] Z^2\)

\(\frac{1}{\lambda} = R_H \left[ \frac{1}{1} - \frac{1}{4} \right] \times 1\)

\(\frac{1}{\lambda} = \frac{3}{4}R_H\)

\(\lambda = \frac{4}{3 R_H}\)

\(\frac{1}{\lambda} = R_H \left[ \frac{1}{\frac{1}{n_1 ^2} - \frac{1}{n_2 ^2}} \right] Z^2\)

\(\frac{1}{\lambda} = R_H \left[ \frac{1}{4} - \frac{1}{9} \right]\)

\(\lambda = \frac{36}{5R_H}\)