Question

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of n + z. (ionisation energy of hydrogen atom = 13.6 eV).

Answer 1

9

Answer 2

Let $E_n = -13.6 \frac{Z^2}{n^2}$ be the energy of an electron in the $n^{th}$ state of a hydrogen-like atom with atomic number Z.

The first excited state corresponds to $n=2$. The second excited state corresponds to $n=3$.

Scenario 1: Transition from state $n$ to the first excited state ($n=2$) by emitting two photons of energies 10.20 eV and 17.00 eV. Total energy emitted, $\Delta E_1 = 10.20 + 17.00 = 27.20$ eV. This energy corresponds to the difference between the initial state $n$ and the final state $n=2$: $E_n - E_2 = -27.20$ eV $E_2 - E_n = 27.20$ eV $(-13.6 \frac{Z^2}{2^2}) - (-13.6 \frac{Z^2}{n^2}) = 27.20$ $13.6 Z^2 \left(\frac{1}{4} - \frac{1}{n^2}\right) = 27.20$ $Z^2 \left(\frac{n^2 - 4}{4n^2}\right) = \frac{27.20}{13.6} = 2$ (Equation 1)

Scenario 2: Transition from the same excited state $n$ to the second excited state ($n=3$) by emitting two photons of energies 4.25 eV and 5.95 eV. Total energy emitted, $\Delta E_2 = 4.25 + 5.95 = 10.20$ eV. This energy corresponds to the difference between the initial state $n$ and the final state $n=3$: $E_3 - E_n = 10.20$ eV $(-13.6 \frac{Z^2}{3^2}) - (-13.6 \frac{Z^2}{n^2}) = 10.20$ $13.6 Z^2 \left(\frac{1}{9} - \frac{1}{n^2}\right) = 10.20$ $Z^2 \left(\frac{n^2 - 9}{9n^2}\right) = \frac{10.20}{13.6} = \frac{3}{4}$ (Equation 2)

Now, divide Equation 1 by Equation 2: $\frac{Z^2 \left(\frac{n^2 - 4}{4n^2}\right)}{Z^2 \left(\frac{n^2 - 9}{9n^2}\right)} = \frac{2}{3/4}$ $\frac{n^2 - 4}{4n^2} \times \frac{9n^2}{n^2 - 9} = \frac{8}{3}$ $\frac{9(n^2 - 4)}{4(n^2 - 9)} = \frac{8}{3}$ $27(n^2 - 4) = 32(n^2 - 9)$ $27n^2 - 108 = 32n^2 - 288$ $32n^2 - 27n^2 = 288 - 108$ $5n^2 = 180$ $n^2 = 36 \implies n = 6$ (since $n$ must be positive)

Substitute $n=6$ into Equation 1: $Z^2 \left(\frac{6^2 - 4}{4 \times 6^2}\right) = 2$ $Z^2 \left(\frac{36 - 4}{4 \times 36}\right) = 2$ $Z^2 \left(\frac{32}{144}\right) = 2$ $Z^2 \left(\frac{2}{9}\right) = 2$ $Z^2 = 9 \implies Z = 3$ (since $Z$ must be positive)

The value of $n + Z = 6 + 3 = 9$.