Question

If proton in H-nucleus be replaced by positron having the same mass as that of proton but same charge as that of proton, then considering the nuclear motion, the wave number of the lowest energy transition of ${\text{H}}{{\text{e}}^ + }$ ion in Lyman series will be equal to:
A.$2{{\text{R}}_{\text{H}}}$
B.${\text{3}}{{\text{R}}_{\text{H}}}$
C.${\text{4}}{{\text{R}}_{\text{H}}}$
D.${{\text{R}}_{\text{H}}}$

Answer 1

The value of Rydberg formula will remain the same for the hydrogen nucleus and the positron. Helium nucleus has 2 protons. The lyman series starts from n value 1.

Formula used:
\dfrac{1}{\lambda } = \mathop \nu \limits^\\_ = {{\text{R}}_{\text{H}}}{{\text{Z}}^2}\left[ {\dfrac{1}{{{\text{n}}_1^2}} - \dfrac{1}{{{\text{n}}_2^2}}} \right]
Here $\lambda$ is wavelength, \mathop \nu \limits^\\_ is wave number, ${{\text{R}}_{\text{H}}}$ is Rydberg constant, Z is nuclear charge, ${{\text{n}}_1}$ is initial energy state and ${{\text{n}}_2}$ is final energy state.

Complete step by step solution:
The atomic number of hydrogen is 1. The nucleus of hydrogen contains only 1 proton; it does not contain any neutron. Hence if we replace the proton by the positron which has the same mass and charge as that of the proton that will make no change in the nucleus of hydrogen. Hence the formula for Rydberg constant which is defined for hydrogen and hydrogen like species will be applicable here too.
Now we need to calculate the value of the wave number for the ${\text{H}}{{\text{e}}^ + }$ ion. The atomic number of helium is 2. It has 2 protons, 2 neutrons in the nucleus and 2 electrons outside. The ${\text{H}}{{\text{e}}^ + }$ ion will contain 2 protons and 2 neutrons but one electron. Hence the electronic configuration which is defined with the number of electrons will remain the same as hydrogen. The rydberg formula will be applicable here too:
The value of Z will be 2 as the nucleus contains 2 protons. For layman series ${{\text{n}}_1}$ is 1 and ${{\text{n}}_2}$ is 2. Hence we will substitute the known variables in the formula and we will get:
\mathop \nu \limits^\\_ = {{\text{R}}_{\text{H}}}{\left( 2 \right)^2}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right] = 3{{\text{R}}_{\text{H}}}

Thus, the correct option is option (B).

Note:
Hydrogen spectra contain five lines named as layman, balmer, paschen, bracket and pfund. The lowest energy level for hydrogen starts from 1 and not from zero, because if it starts from zero the zero point energy will be zero and this is not possible.