Question

The shortest wavelength in the Lyman series of hydrogen spectrum is 912 Å corresponding to a photon energy of 13.6 eV. The shortest wavelength in the Balmer series is-

• A. 3648 Å
• B. 228 Å
• C. 6566 Å
• D. 8208 Å

Answer 1

Answer: (A)

3648 Å

Answer 2

$\frac{1}{\lambda}$ = R Z² $\left\lbrack \frac{1}{n_{1}^{2}}–\frac{1}{n_{2}^{2}} \right\rbrack$

$\frac{1}{912}$ = R × 1 × $\left\lbrack \frac{1}{1}–\frac{1}{\infty} \right\rbrack$

$\frac{1}{\lambda}$ = R × 1 × $\left\lbrack \frac{1}{4}–\frac{1}{\infty} \right\rbrack$

$\frac{\lambda}{912}$ = 4

l = 912 × 4 = 3648 Å