Question

Energy of $4^{th}$ Bohr orbit of $H{e}^{+}$ is x, then calculate the energy of $4^{th}$ excited state of $L{i}^{2+}$ in terms of x.

• A. $\frac{25}{36}x$
• B. $\frac{36}{25}x$
• C. $\frac{18}{25}x$
• D. $\frac{25}{18}x$

Answer 1

The correct option is B $\frac{36}{25}x$

$ForH{e}^{+}z=2,n=4$
$x=-13.6\times \frac{4}{16}\cdots \left(i\right)$
$ForL{i}^{2+}z=3,n=5\left(4^{th}excitedstate\right)$
$E_s=-13.6\times \frac{9}{25}\cdots \left(ii\right)$
Divide equation (i) by (ii)
$\frac{x}{E_s}=\frac{4}{16}\times \frac{25}{9}$
$E_s=\frac{36}{25}x$