The wavelength of the radiation emitted is λ _ 0 when an electron...

_____n = 4$</p><p>$_____n = 3$</p><p>$_____n = 2$</p><p>$_____n = 1$</p><p>The wavelength of the light released when an electron jumps from higher energy state$\left({n}_{2}\right)$to lower energy state \left({n}_{1}\right)$ is $\dfrac{1}{\lambda} = R\left(1{n}_{1}^{2} - \dfrac{1}{{n}_{2}^{2}}\right)$, here, $\lambda$ is wavelength, $R$ is Rydberg Constant.

When transition is from second excited state to first excited state,
$n = 3 \to n = 2$

$\dfrac{1}{{\lambda}_{0}} = R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right) . . .$\left(1\right)$</p><p>When transition is from third excited state to second orbit</p><p>$n = 4 \to n = 2$</p><p>$\dfrac{1}{\left(\dfrac{20}{x}{\lambda}_{0}\right)} = R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{4}^{2}}\right) . . . \left(2\right)$</p><p>Dividing \left(2\right)by\left(1\right)$

$⇒ \dfrac{x}{20} = \dfrac{\dfrac{1}{{2}^{2}} - \dfrac{1}{{4}^{2}}}{\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}} = \dfrac{\dfrac{1}{4} - \dfrac{1}{16}}{\dfrac{1}{4} - \dfrac{1}{9}} = \dfrac{\dfrac{3}{16}}{\dfrac{5}{36}} = \dfrac{27}{20}$

$⇒ x = 27$