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)
$\frac{x}{20}=\frac{\frac{1}{2^2}-\frac{1}{4^2}}{\frac{1}{2^2}-\frac{1}{3^2}}$
$x=27$

Answer 2

1. Transition from second excited state (\(n=3\)) to first excited state (\(n=2\)): \[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right), \] where \(R\) is the Rydberg constant.
Simplify: \[ \frac{1}{\lambda_0} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36}. \]
2. Transition from third excited state (\(n=4\)) to second orbit (\(n=2\)): \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right). \] Simplify: \[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = R \cdot \frac{3}{16}. \]
3. Wavelength ratio: \[ \frac{\lambda_0}{\lambda} = \frac{\frac{5}{36}}{\frac{3}{16}} = \frac{5 \cdot 16}{36 \cdot 3} = \frac{80}{108} = \frac{20}{27}. \] Thus: \[ \lambda = \frac{20}{27} \lambda_0. \]
Hence, \(x = 27\). The wavelength of radiation depends on the energy difference between the two levels. Simplify Rydberg's formula step by step to calculate ratios.