Question

It requires $68eV$ of energy to excite the electron from the second Bohr orbit to the third Bohr orbit in a particular single electron species having a nuclear charge of $+Ze$ , where $Z$ is a constant and $e$ is the magnitude of the electronic charge. Figure out the value of $Z$ .

Answer 1

The energy required to excite the electron from second (n=2) Bohr orbit to the third (n=3) Bohr orbit is given by
$\Delta E=E_{3}-E_{2}$
$=\left(- \frac{Z^{2} \times 13 . 6}{3^{2}}\right)-\left(- \frac{Z^{2} \times 13 . 6}{2^{2}}\right)$
$=z^{2}\times 13.6\left(\frac{1}{4} - \frac{1}{9}\right)$
$=Z^{2}\times 13.6\times \frac{5}{36}\ldots .\left(\right.i\left.\right)$
$\text{ But }ΔE=E_{3}-E_{2}=68eV$
From equations (i) and (ii),
$\therefore 68eV=Z^{2}\times 13.6\times \frac{5}{36}eV$
$\therefore Z^{2}=\frac{36}{5}\times \frac{68}{13 . 6}=36\Rightarrow Z=6$