Question
In which of the following transitions will the wavelength be minimum?
✓
Answer
- n = 2 to n = 1
For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}\Big)$
Here, R is the Rydberg constant.
For the transition from n = 5 to n = 4, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{4^2}-\frac{1}{5^2}\Big)$
$\lambda=\frac{400}{9\text{RZ}^2}$
For the transition from n = 4 to n = 3, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{3^2}-\frac{1}{4^2}\Big)$
$\lambda=\frac{144}{7\text{RZ}^2}$
For the transition from n = 3 to n = 2, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$
$\lambda=\frac{36}{5\text{RZ}^2}$
For the transition form n = 2 to n = 1, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda=\frac{2}{\text{RZ}^2}$
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n = 2 to n = 1 will be minimum.
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