Question

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

• A. $4000\mathring{A}$
• B. $100\mathring{A}$
• C. $40\mathring{A}$
• D. $10\mathring{A}$

Answer 1

Answer: (D)

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,
$Z=11,\frac{1}{{\lambda }^{\prime }}=121R\frac{3}{4}$
$\Rightarrow \frac{{\lambda }^{\prime }}{\lambda }=\frac{3R}{4}\times \frac{4}{121R\times 3}$
$\Rightarrow {\lambda }^{\prime }=\frac{\lambda }{121}$
$=\frac{1210}{121}=10\mathring{A}$