Question

A hydrogen like atom with atomic number 'Z' is in higher excited state of quantum number 'n'. This excited state atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17.0 eV respectivley. Alternatively, the atom from the same excited state can make a transition to the 2nd excited state by emitting two photons of energies 4.25 eV and 5.95 eV respectively. Calculate the value of 'Z'.

Answer 1

3

Answer 2

The energy of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2}$ eV.

Transition to the first excited state ($n=2$): Total energy emitted = $10.2 \, \text{eV} + 17.0 \, \text{eV} = 27.2 \, \text{eV}$. Equation: $E_n - E_2 = 27.2 \, \text{eV} \implies 13.6 Z^2 \left(\frac{1}{2^2} - \frac{1}{n^2}\right) = 27.2 \implies Z^2 \left(\frac{1}{4} - \frac{1}{n^2}\right) = 2$.

Transition to the second excited state ($n=3$): Total energy emitted = $4.25 \, \text{eV} + 5.95 \, \text{eV} = 10.2 \, \text{eV}$. Equation: $E_n - E_3 = 10.2 \, \text{eV} \implies 13.6 Z^2 \left(\frac{1}{3^2} - \frac{1}{n^2}\right) = 10.2 \implies Z^2 \left(\frac{1}{9} - \frac{1}{n^2}\right) = \frac{10.2}{13.6} = \frac{3}{4}$.

Dividing the two equations: $\frac{Z^2 (\frac{1}{4} - \frac{1}{n^2})}{Z^2 (\frac{1}{9} - \frac{1}{n^2})} = \frac{2}{3/4} = \frac{8}{3}$ $\frac{\frac{n^2-4}{4n^2}}{\frac{n^2-9}{9n^2}} = \frac{8}{3}$ $\frac{9(n^2-4)}{4(n^2-9)} = \frac{8}{3}$ $27(n^2-4) = 32(n^2-9)$ $27n^2 - 108 = 32n^2 - 288$ $5n^2 = 180 \implies n^2 = 36 \implies n = 6$.

Substitute $n=6$ into the first equation: $Z^2 \left(\frac{1}{4} - \frac{1}{36}\right) = 2$ $Z^2 \left(\frac{9-1}{36}\right) = 2$ $Z^2 \left(\frac{8}{36}\right) = 2$ $Z^2 \left(\frac{2}{9}\right) = 2 \implies Z^2 = 9 \implies Z = 3$.