Atoms · Question 17

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An electron in the hydrogen atom initially in the fourth excited state makes a transition to nth energy state by emitting a photon of energy 2.86 eV. The integer value of n will be .

Answer

Solution

Fourth excited state ni = 5

E5 = - 13.652 = -0.544 eV

Energy of the final state: En = E5 - 2.86 = -0.544 - 2.86 = -3.404 eV

- 13.6n2 = -3.404

n2 = 13.63.404 4

n = 2

Answer: 2

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