Question

The ground state energy of the electron in the hydrogen atom is equal to:
A. The ground state energy of the electron in $H{{e}^{+}}$
B. The first excited state energy of the electron in $H{{e}^{+}}$
C. The first excited state energy of the electron in $L{{i}^{+2}}$
D. The ground state energy of the electron in $B{{e}^{+3}}$

Answer 1

This is a direct question. The formula for computing the energy level of the electrons should be used to solve this problem. The ratio of the atomic number to the principal quantum number for the ground state hydrogen atom equals the ratio of the atomic number to the principal quantum number for the first excited state of the $H{{e}^{+}}$. Thus, using this, we will solve the problem.
Formula used:
${{E}_{n}}=-\dfrac{R{{Z}^{2}}}{{{n}^{2}}}$

Complete step-by-step solution
The formula for computing the energy level of the electrons should be used to solve this problem.
The formula that defines the relation between the Rydberg constant, the atomic number, and the principal quantum number is given as follows.
${{E}_{n}}=-\dfrac{R{{Z}^{2}}}{{{n}^{2}}}$
Where R represents the Rydberg constant, Z is the atomic number and n is the principal quantum number.
Firstly, we will compute the value of the ground state energy of the electron in the hydrogen atom. So, we get,
The atomic number of the ground state hydrogen atom is 1 and the principal quantum number of the ground state is 1.

& {{E}_{H}}=-\dfrac{R\times {{1}^{2}}}{{{1}^{2}}} \\\ & {{E}_{H}}=-R \\\ \end{aligned}$$ Now, we will compute the value of the first excited state energy of the electron in $$H{{e}^{+}}$$. So, we get, The atomic number of the first excited state$$H{{e}^{+}}$$ is 2 and the principal quantum number of the first excited state is 2. $$\begin{aligned} & {{E}_{H{{e}^{+}}}}=-\dfrac{R\times {{2}^{2}}}{{{2}^{2}}} \\\ & {{E}_{H{{e}^{+}}}}=-R \\\ \end{aligned}$$ Upon comparing the values of both the equations, we get that, The first excited state energy of the electron in $$H{{e}^{+}}$$ equals the ground state energy of the electron in the hydrogen atom. $$\therefore $$ The first excited state energy of the electron in $$H{{e}^{+}}$$ equals the ground state energy of the electron in the hydrogen atom. **As, the value of the ground state energy of the electron in the hydrogen atom equals the first excited state energy of the electron in $$H{{e}^{+}}$$, thus, the option (B) is correct.** **Note:** Knowing the formulae is the most important thing while solving such problems. At least the values of the atomic number and the number of molecules of the atomic gases should be known to solve this type of problem.