Question

If the Rydberg Constant for hydrogen atom is $R$, then the Rydberg constant for positronium atom is

& A.2R \\\ & B.R \\\ & C.\dfrac{R}{2} \\\ & D.4R \\\ \end{aligned}$$

Answer 1

Hint: Rydberg constant of any atom is given by $R=\dfrac{{{m}_{e}}{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{3}}c}$. Here everything is constant except mass which may vary. Rydberg constant is directly proportional to mass of electron ( which may vary), this will help us to find Rydberg constant for positronium atom.

Complete step by step answer:
A hydrogen atom which has atomic number one , has one electron and one proton. A Hydrogen atom has a nucleus in which one proton resides.
According to Bohr Postulates electrons revolve around the nucleus of the hydrogen atom.
Mass of the electron is represented by ${{m}_{e}}$ and charge is represented by $e$.
$e=1.6\times {{10}^{-19}}C$
${{m}_{e}}=9.1\times {{10}^{-31}}kg$
Rydberg Constant: When we calculate the wave numbers of atomic spectra by the atom. Which is caused when an electron jumps from lower shell to higher shell or from higher shell to lower shell. Then we get Rydberg constant in the formula.
Value of Rydberg Constant is equals to $=1.097\times {{10}^{7}}{{m}^{-1}}$
Positronium Atom: A positronium atom is the one without nucleus, it has one electron and one antiparticle of electron known as positron. Which revolve around themselves.
Mass of an antiparticle of an electron is same as the mass of electron i.e. ${{m}_{e}}$.
But the reduced mass of an atom is given by the formula:
$\mu =\dfrac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}$
Where:
${{m}_{1}}={{m}_{2}}={{m}_{e}}$
Therefore the reduced mass of positronium atom:

& \mu =\dfrac{{{m}_{e}}{{m}_{e}}}{{{m}_{e}}+{{m}_{e}}} \\\ & \mu =\dfrac{{{m}_{e}}}{2} \\\ \end{aligned}$$ And we know that the Rydberg Constant is given by: $$R=\dfrac{{{m}_{e}}{{e}^{4}}}{8{{\varepsilon }_{0}}^{2}{{h}^{3}}c}$$ By this we can easily see Rydberg constant is directly proportional to mass of electron $$R\alpha {{m}_{e}}$$ Here mass of positronium atom is $$\dfrac{{{m}_{e}}}{2}$$ $${{R}_{1}}=\dfrac{R}{2}$$ Where $${{R}_{1}}$$ is the Rydberg constant for a Positronium atom. Thus Rydberg constant for positronium atom is equal to half the Rydberg constant of hydrogen atom. Answer is option C. $$\dfrac{R}{2}$$ Note: Rydberg constant has the value $$1.097\times {{10}^{7}}{{m}^{-1}}$$. But we have seen in this question the value of the Rydberg constant becomes half of this value. Students must be careful because they think the value of constant always remains the same. So go through the formula and check each factor which may vary.