Question

The wavelength of the radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of hydrogen atom If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$ The value of $x$ is

Answer 1

Correct Answer: 27

The correct answer is 27
________$n=4$
_________$n=3$
_________ $n=2$
__________$n=1$
Second excited state $\to$ first excited state
$n=3\to n=2$
$\frac{hc}{{\lambda }_0}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right).....$(i)
Third excited state $\to$ second orbit $n=4\to n=2$
$\frac{hc}{\left(20{\lambda }_0/x\right)}=13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right)......$(ii)
(ii) $÷$ (i)
$\frac{x}{20}=\frac{\frac{1}{2^2}-\frac{1}{4^2}}{\frac{1}{2^2}-\frac{1}{3^2}}$
$x=27$