Question

The ground state energy of a hydrogen atom is -13.6eV. The energy of second excited state of $H{e^ + }$ ion in eV is:
A. -27.2
B. -3.4
C. -54.4
D. -6.04

Answer 1

Hint: State energy of an energy level is a product of ground state energy and square of the ratio of nuclear charge to the number of the energy level. Second excited state is the third energy level. The nuclear charge is the number of protons in the nucleus which is 2 for He.

Complete answer:
Based on the Bohr model, an atom electron can take quantised energy states and it can change its energy state when it absorbs energy of exact quantity equal to the difference between the energy states. Here we have been provided with the ground state energy of the hydrogen atom which is ${E_1} = - 13.6eV$. The subscript 1 shows that it is the lowest energy level known as the ground state and the negative sign is because that much energy the electron will absorb to escape the nucleus. Energy of ${n^{th}}$ energy level or ${(n - 1)^{th}}$ excited state is given by: ${E_n} = \dfrac{{{E_1}}}{{{n^2}}}$
This energy increases proportionally as squared of the nuclear charge, or number of protons in the nucleus. The expression is as follows: ${E_n} = {E_1}\dfrac{{{Z^2}}}{{{n^2}}}$, where $Z$ is the number of protons in the nucleus.
For Helium the number of protons is equal to 2. The question asks for the energy of the second excited state or the third energy level. Substituting, $n = 3$ in the above expression, we get,
${E_3} = {E_1}\dfrac{{{2^2}}}{{{3^2}}} = - 13.6 \times \dfrac{4}{9} = - 6.04eV$. After comparing this result with the options, we can say that option D is correct.

Note: You will have to be careful whether they are asking the energy level or the excited state. If they are asking excited state then numbering starts with the second energy level as first excited state. So in the above problem, we had a third energy level as our second excited state.