Physics - Modern Physics Question with Solution | TestHub

The energy of an electron in the $n$th orbit of a hydrogen-like atom is given by $E_n=-13.6\frac{Z^2}{n^2}$ eV, where $Z$ is the atomic number. The energy required to excite the electron from the second to the third orbit is:

$\Delta E=E_3-E_2=-13.6Z^2(\frac{1}{3^2}-\frac{1}{2^2}{)}^{}$

$-13.6Z^2(\frac{1}{9}-\frac{1}{4})=-13.6Z^2(\frac{4-9}{36})=\frac{13.6\times 5}{36}{Z}^2$

Given $\Delta E=47.2$ eV,

we have:

$47.2=\frac{13.6\times 5}{36}{Z}^2$

$Z^2=\frac{47.2\times 36}{13.6\times 5}=\frac{47.2\times 36}{68}=\frac{1699.2}{68}=24.98\approx 25$

$Z=\sqrt{25}=5$