Question

The first line of the Lyman series in a hydrogen spectrum has a wavelength of $1210 ??. The corresponding line of a hydrogen- like atom of$Z = 11$ is equal to

• A. $4000 ??
• B. $100 ??
• C. $40 ??
• D. $10 ??

Answer 1

Answer: (D)

$10 ??

Answer 2

For first line of Lyman series $n_1 = 1, n_2 = 2$ $\therefore \frac{1}{\lambda} = Z^2 R \frac{3}{4}$ In the case of hydrogen atom, $Z = 1, \frac{1}{\lambda} = R{\frac{3}{4}}$ For hydrogen-like atom, $Z = 11, \frac{1}{\lambda'} = 121 R\frac{3}{4}$ $\Rightarrow \frac{\lambda'}{\lambda} = \frac{3R}{4} \times \frac{4}{121R \times 3}$ $\Rightarrow \lambda' = \frac{\lambda}{121}$ $= \frac{1210}{121} = 10,??