Question

The wavelength of the first line of Lyman series for the hydrogen atom is equal to that of the second line of the Balmer series for hydrogen-like ions. The atomic number Z of hydrogen-like ion is
A. 3
B. 4
C. 1
D. 2

Answer 1

The atomic number of an element is the number of protons found in the nucleus of every atom of that element.
The Balmer series is a part of atomic physics which is a set of six named series describing the spectral line emission of hydrogen atoms calculated by using the Balmer formula. Balmer series of hydrogen lines the Balmer series of atomic hydrogen. These lines are emitted when electrons of hydrogen atom transitions from the n=3.
Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines from the hydrogen atom as electrons go from $n \geqslant 0$ to $n = 1$, the lowest energy level of the electrons.
In this question it is given that the wavelength of the first line of Lyman series for the hydrogen atom is equal to that of the second line of the Balmer series for hydrogen, then find the atomic number by equating the formula.

Complete step by step answer:
The first line of the Lyman series for the hydrogen atom is given as
$\dfrac{1}{{{\lambda _{Lyman}}}} = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
The atom of hydrogen is transiting from line 1 to line 2
${n_1} = 1$
${n_2} = 2$
So we can write

$$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\\ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right) \\\ = R\left( {\dfrac{1}{1} - \dfrac{1}{4}} \right) \\\ = R\left( {\dfrac{{4 - 1}}{4}} \right) \\\ \dfrac{1}{\lambda } = \dfrac{3}{4}R - - - (i) \\\$$

Now the second line of the Balmer series for the hydrogen atom is given as
$\dfrac{1}{{{\lambda _{Balmer}}}} = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
The atom of hydrogen is transiting from line 2 to line 4
${n_1} = 2$
${n_2} = 4$
By putting

$$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\\ \dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right) \\\ = R{Z^2}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) \\\ = R{Z^2}\left( {\dfrac{{4 - 1}}{{16}}} \right) \\\ \dfrac{1}{\lambda } = \dfrac{3}{{16}}{Z^2}R - - - (ii) \\\$$

Since the wavelength of the first line of Lyman series for the hydrogen atom is equal to that of the second line of Balmer series, so we can write equation (i) = equation (ii), therefore

$$\dfrac{3}{4}R = \dfrac{3}{{16}}{Z^2}R \\\ {Z^2} = 4 \\\ Z = 2 \\\$$

Hence the atomic number Z of the hydrogen atom$= 2$

So, the correct answer is “Option D”.

Note:
Students should not get confused with the levels of the atoms shifting, the term $\dfrac{1}{{{\lambda _{Lyman}}}} = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$ should always be positive when the atoms shifts from lower level to the higher level whereas $\dfrac{1}{{{\lambda _{Lyman}}}} = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$ will be negative when the atoms shifts from higher level to the lower level.