Question

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

• A. 4000 Å
• B. 100 Å
• C. 40 Å
• D. 10 Å

Answer 1

Answer: (D)

10 Å

Answer 2

By Bohr’s formula

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

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

For hydrogen like atom $Z = 11$

$\frac{1}{\lambda'} = 121R\frac{3}{4} \Rightarrow \frac{\lambda'}{\lambda;} = \frac{3R}{4} \times \frac{4}{121R \times 3} = \frac{1}{121}$

$\lambda' = \frac{\lambda}{121} = \frac{1210}{121} = 10Å$