Question 430: Atoms & Nuclei

Question

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

Accepted Answer

According to Rydberg's formula,
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
When electron jumps from $2^{	ext {nd }}$ excited state to first excited state,
$\mathrm{n}_2=3, \mathrm{n}_1=2$, we get
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}ight) $$
When electron jumps from $3^{	ext {rd }}$ exited state to $2^{	ext {nd }}$ orbit, $\mathrm{n}_2=4, \mathrm{n}_1=2, \lambda=\lambda_0$, we get:
$$ \begin{array}{rlrl}  & & \frac{1}{\lambda_0} & =R\left(\frac{1}{4^2}-\frac{1}{2^2}ight) \ & 	herefore & \frac{\lambda_0}{\lambda} & =\frac{R\left(\frac{1}{2^2}-\frac{1}{3^2}ight)}{R\left(\frac{1}{2^2}-\frac{1}{4^2}ight)}=\frac{5}{36} 	imes \frac{16}{3}=\frac{20}{27} \ & 	herefore & \lambda_0 & =\frac{20}{27} \lambda \ & \Rightarrow & x=27 \end{array} $$

Answer

18

Answer

27

Answer

21

Answer

36

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