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Question
A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?
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Solution
Energy (E) of the nth state of hydrogen atom is given by
`E =13.6/n^2 eV`
For n = 6,
`therefore E = (-13.6)/36 = -0.377777777 eV`
Energy of hydrogen atom in the ground state = −13.6 eV
Energy emitted in the second transition = Energy of ground state + (Energy of hydrogen atom in the 6th state + Energy of photon)
= 13.6 - (0.37777 + 1.13)
= 12.09 = 12.1 eV
(b)
Energy in the intermediate state = (Energy of photon emitted in the first transition) + (Energy of n = 6 state
= 1.13 eV + 0.377 eV
= 1.507 eV
Energy of the nth state can expressed as
`13.6/n^2 = 1.507`
`therefore n = sqrt((13.6)/1.507) = sqrt(9.024) ≈3`
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