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Question
In which of the following transitions will the wavelength be minimum?
Options
n = 5 to n = 4
n = 4 to n = 3
n = 3 to n = 2
n = 2 to n = 1
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Solution
n = 2 to n = 1
For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by
`1/lamda = RZ^2 (1/n^1 - 1/n^2)`
Here, R is the Rydberg constant.
For the transition from n = 5 to n = 4, the wavelength is given by
`1/lamda = RZ^2 (1/4^2 - 1/5^2)`
`lamda = 400/(9RZ^2)`
For the transition from n = 4 to n = 3, the wavelength is given by
`1/lamda = RZ^2 (1/3^2 - 1/4^2)`
`lamda = (144)/ (7RZ^2)`
For the transition from n = 3 to n = 2, the wavelength is given by
`1/lamda = RZ^2 (1/2^2 - 1/3^2 )`
`lamda = (36)/(5RZ^2)`
For the transition from n = 2 to n = 1, the wavelength is given by
`1/lamda = RZ^2 (1/2^2 - 1/3^2)`
`lamda = 2/(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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