Question

The ratio of the maximum to the minimum wavelengths of an electron that emits radiation is 1.8. What energy level (n) will its first excited state be at?

Answer 1

We know,
$\lambda \propto \frac{1}{E}$

When an electron transits from $\infty$ to current level $x$.
$\frac{1}{{\lambda }_{min}}=\frac{1}{R_H}(\frac{1}{x^2}-\frac{1}{\infty })...\left(i\right)$

When an electron transits from $(x+1)$ to current level $x$.
$\frac{1}{{\lambda }_{max}}=\frac{1}{R_H}(\frac{1}{x^2}-\frac{1}{(x+1{)}^2})...\left(ii\right)$

Dividing $e{q}^n\left(i\right)$ by $e{q}^n\left(ii\right)$, we get
$\frac{{\lambda }_{max}}{{\lambda }_{min}}=1.8$

$\Rightarrow \frac{1}{x^2}\times \frac{1}{(\frac{1}{x^2}-\frac{1}{(x+1{)}^2})}=1.8$

$\Rightarrow \frac{(x+1{)}^2}{2x+1}=1.8$

$\Rightarrow x^2-1.6x-0.8=0$
$\Rightarrow (x-2)(5x-2)=0$
$x=2$ (+ve value)
So 1st excited state will be: $x+1=2+1=3$