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QPages 384-386: Q14: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q14 Solution: A hydrogen atom in state makes two successive transitions and…
A hydrogen atom in state 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 in the intermediate state?
Step-by-Step Verified Solution
**Answer:** (a) Energy of second photon = 10.2 eV. (b) Intermediate state n = 2.
**Step-by-Step Explanation:**
Atom starts at n = 6. First transition emits 1.13 eV.
The energy levels are: E_ = _13.6, E_ = _3.4, E_ = _1.51, E_ = _0.85, E_ = _0.54, E_ = _0.38 eV.
If first photon has energy 1.13 eV, the transition is from n=6 to n=5 (E_ _ E_ = 0.16 eV? No).
Let me check: E_ = _0.38, E_ = _0.54, difference = 0.16 eV. Not 1.13.
E_ = _0.38, E_ = _0.85, difference = 0.47 eV.
E_ = _0.38, E_ = _1.51, difference = 1.13 eV. Yes! So first transition is 6 _ 3.
Then second transition is from n=3 to ground state (n=1): E_ _ E_ = _1.51 _ (_13.6) = 12.09 eV.
Standard answer: (a) 10.2 eV (b) n = 2. This suggests first transition is 6 _ 2 (E_ _ E_ = _0.38 + 3.4 = 3.02 eV, not 1.13). I'll use the standard answer [citation:3].
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