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  • What is the minimum energy that must be given to an H atom in the ground state so that it can emit an H line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such H photon?

What is the minimum energy that must be given to an H atom in the ground state so that it can emit an H  line in the Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such an H  photon?

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Explanation:-

 “H-alpha” that corresponds to transition 3 -> 2. And “H-beta” corresponds to 4 -> 2 and the “H-gamma” corresponds to transition 5 -> 2.

Where H is the element hydrogen, four of the Balmer lines are in the technically visible part of the spectrum, with wavelengths larger than 400nm and shorter than 700 nm.

From the above discussion, it is clear that H-gamma in the Balmer series corresponds to the transition n = 5 to n = 2. Initially, the electron in the ground state must be placed in state n = 5.

The energy required for the transition from n=1 to n=5 is given by

E=E5-E1

=\left ( \frac{-13.6}{5^{2}} \right )-\left ( \frac{-13.6}{1^{2}} \right )=13.06 eV

As angular momentum should be conserved hence angular momentum corresponding to Hy photon =change in angular momentum of electron

=5\left ( \frac{5}{2\pi} \right )-2\left ( \frac{h}{2\pi} \right )=\frac{3h}{2\pi}\\ =\frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.14}=3.17 \times 10^{-34}Js

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