Question

An electron revolves round a nucleus of charge $+Ze$. If the energy required to excite the electron from second to third Bohr orbit is $47.2eV$, then the value of $Z$ will be

• A. $5$
• B. $4$
• C. $3$
• D. $1$

Answer 1

The correct option is A $5$

Energy needed to ionise an atom from $n^{th}$ orbit with charge $Z=13.6\frac{z^2}{n^2}eV$.
$\Rightarrow 13.6z^2\left(\frac{1}{2^2-\frac{1}{3^2}}\right)=47.2$
$\Rightarrow \frac{13.6}{36}\times 5Z^{2}47.0$

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

$\Rightarrow Z^2=25$

$\Rightarrow Z=5$