Question: 74. The energy of an electron in the ground state ( \mathrm{n}=1 ) for \mathrm{He}^{+} ion is -xJ , then that for an electron in \mathrm{n}=2 state for \mathrm{Be}^{3+} ion in J is:
(1) -x
(2) -\frac{x}{9}
(3) -4x
(4) -\frac{4}{9}x
Answer: Option (1)
Explanation:
For a hydrogen-like ion, the energy of an electron in the nth orbit is given by:
E_n = -13.6 \frac{Z^2}{n^2} \ \text{eV}For the \mathrm{He}^{+} ion, the atomic number is Z=2.
In the ground state n=1, the energy is given as:
E_1 = -13.6 \frac{(2)^2}{(1)^2} = -54.4 \ \text{eV}This energy is given as -x \ \text{J}.
For the \mathrm{Be}^{3+} ion, the atomic number is Z=4.
In the n=2 state, the energy is:
E_2 = -13.6 \frac{(4)^2}{(2)^2} = -13.6 \times \frac{16}{4} = -54.4 \ \text{eV}The magnitude of energy is the same as that of the \mathrm{He}^{+} ion in the ground state.
Therefore, the energy in joules is also:
E = -x \ \text{J}Hence, the correct answer is Option (1).