Question

For hydrogen atom, Rydberg constant $\left( {{{\text{R}}_H}} \right)$ is ${\text{x }}{{\text{m}}^{{\text{ - 1}}}}$. Then for $H{e^ + }$ ion, the corresponding value of this constant will be:
(a) ${\text{x c}}{{\text{m}}^{{\text{ - 1}}}}$
(b) ${\text{400 x c}}{{\text{m}}^{{\text{ - 1}}}}$
(c) ${\text{4 x c}}{{\text{m}}^{{\text{ - 1}}}}$
(d) ${\text{0}} \cdot {\text{04 x c}}{{\text{m}}^{{\text{ - 1}}}}$

Answer 1

Hint:Rydberg constant is denoted by \[{R_\infty }\] for heavier atoms, and \[{R_H}\] for hydrogen. For dealing with problems related to hydrogen, we use the following formula: \[\dfrac{1}{\lambda } = {R_H}{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)\], where,
$\lambda $ is the wavelength of the electron, and \[\dfrac{1}{\lambda }\] is the wavenumber,
$Z$ is the atomic number of the atom considered,
${{\text{R}}_H}$ is the Rydberg constant for hydrogen atom,
${n_1}^1{\text{ and }}{n_2}^2$ are the energy states such that ${n_1}{\text{ > }}{n_2}$.

Complete step by step answer: The Rydberg constant is a representation of the limiting value of the highest wave per unit area of any photon that can be emitted from the hydrogen atom. This will tell the wavenumber of the photon which is capable to ionise the hydrogen atom from its ground state.
Step (1):
For hydrogen atom, Rydberg constant $\left( {{{\text{R}}_H}} \right)$ is given as ${\text{x }}{{\text{m}}^{{\text{ - 1}}}}$. We have to find the value of , Rydberg constant for $H{e^ + }$. For this, we will find a relation between the Rydberg constants of both ${\text{H and H}}{{\text{e}}^ + }$.
Step (2):
For hydrogen atom, Rydberg constant equation will be: \[\dfrac{1}{\lambda } = {R_H}{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{ }}{{\text{m}}^{ - 1}}........{\text{(1)}}\]
And for Helium atom, $H{e^ + }$, Rydberg constant equation will be: \[\dfrac{1}{\lambda } = {R_{H{e^ + }}}{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{ c}}{{\text{m}}^{ - 1}}........{\text{(2)}}\]
${{\text{R}}_H}$ is given as ${\text{x }}{{\text{m}}^{{\text{ - 1}}}}$, which is meters, and we know that $1{\text{ m = 100 cm}}$, so, converting the given value for centimetres, we get, ${{\text{R}}_H} = 0 \cdot 01{\text{ x c}}{{\text{m}}^{ - 1}}$. Also, the atomic number of hydrogen atom, and Helium atom are ${\text{1 and 2}}$ respectively. Now, putting all the given values in the equation${\text{(1)}}$, we get,
\[ \Rightarrow \dfrac{1}{\lambda } = 0 \cdot 01{\text{ x c}}{{\text{m}}^{ - 1}}{\left( 1 \right)^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{ c}}{{\text{m}}^{ - 1}}\]
\[ \Rightarrow \dfrac{1}{\lambda } = 0 \cdot 01{\text{ x c}}{{\text{m}}^{ - 1}}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{ c}}{{\text{m}}^{ - 1}}.......{\text{(3)}}\]
Then, putting all the given values in the equation${\text{(2)}}$, we get,
\[ \Rightarrow \dfrac{1}{\lambda } = {R_{H{e^ + }}}{\left( 2 \right)^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{ c}}{{\text{m}}^{ - 1}}\]
\[ \Rightarrow \dfrac{1}{\lambda } = {R_{H{e^ + }}} \times 4\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{ c}}{{\text{m}}^{ - 1}}........{\text{(4)}}\,\]
Dividing equation ${\text{(4) by (3)}}$, we get,
$ \Rightarrow {R_{H{e^ + }}} = \dfrac{4}{{{R_H}}}{\text{ c}}{{\text{m}}^{ - 1}}$
$ \Rightarrow {R_{H{e^ + }}} = \dfrac{4}{{0 \cdot 01}}{\text{c}}{{\text{m}}^{ - 1}}$
$ \Rightarrow {R_{H{e^ + }}} = 0 \cdot 04{\text{ c}}{{\text{m}}^{ - 1}}$

Hence, option (d) is correct, the value of this constant is $0 \cdot 04{\text{ c}}{{\text{m}}^{ - 1}}$.

Note:Rydberg constant is used only with hydrogen-like atoms. Hydrogen like atoms means any species which has only one electron. For example, in the above question Helium in normal has two electrons, but to make it hydrogen like, it loses one electron to form $H{e^ + }$.