In which of the following transitions will the wavelength be minimum?

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KollegeApply · Accepted Answer

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 ($\frac{1}{4}$^2 - $\frac{1}{5}$^2) lamda = 400/(9RZ^2)` For the transition from n = 4 to n = 3, the wavelength is given by `1/lamda = RZ^2 ($\frac{1}{3}$^2 - $\frac{1}{4}$^2) lamda = (144)/ (7RZ^2)` For the transition from n = 3 to n = 2, the wavelength is given by `1/lamda = RZ^2 ($\frac{1}{2}$^2 - $\frac{1}{3}$^2 ) lamda = (36)/(5RZ^2)` For the transition from n = 2 to n = 1, the wavelength is given by `1/lamda = RZ^2 ($\frac{1}{2}$^2 - $\frac{1}{3}$^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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