Question 47: Atoms & Nuclei

Question

The wavelength of the first line of Lyman series for $$\mathrm{H}$$ - atom is equal to that of the second line of Balmer series for a $$\mathrm{H}$$-like ion. The atomic number $$\mathrm{Z}$$ of $$\mathrm{H}$$-like ion is

Accepted Answer

Lyman series of $$\mathrm{H}$$-atom, we can write
$$\frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}ight)$$
where, symbols have their usual meaning and for second line of Balmer series of $$\mathrm{H}$$-like ion
$$\frac{h c}{\lambda}=Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}ight)$$
Therefore, $$\left(\frac{1}{1^2}-\frac{1}{2^2}ight)=Z^2\left(\frac{1}{4}-\frac{1}{16}ight)$$
$$\begin{aligned}
\left(1-\frac{1}{4}ight) & =Z^2\left(\frac{1}{4}-\frac{1}{16}ight) \
\Rightarrow \quad Z & =2.19 \simeq 2
\end{aligned}$$
The approximately value of $$Z$$ will be 2.

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Atoms & Nuclei

The wavelength of the first line of Lyman series for $\mathrm{H} - atom is equal to that of the second line of Balmer series for a \mathrm{H}-like ion. The atomic number \mathrm{Z} of \mathrm{H}$-like ion is

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