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The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
A
2
B
3
C
4
D
1
Detailed Solution
Rydberg formula: $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
First line of Lyman series of hydrogen (Z = 1, $n_1 = 1$, $n_2 = 2$): $\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$
Second line of Balmer series of the ion ($n_1 = 2$, $n_2 = 4$): $\frac{1}{\lambda} = RZ^2\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3RZ^2}{16}$
The wavelengths are equal: $\frac{3R}{4} = \frac{3RZ^2}{16}$
$Z^2 = 4$
Z = 2
First line of Lyman series of hydrogen (Z = 1, $n_1 = 1$, $n_2 = 2$): $\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$
Second line of Balmer series of the ion ($n_1 = 2$, $n_2 = 4$): $\frac{1}{\lambda} = RZ^2\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3RZ^2}{16}$
The wavelengths are equal: $\frac{3R}{4} = \frac{3RZ^2}{16}$
$Z^2 = 4$
Z = 2