lf an electron jumps from the second orbit to first orbit in hydrogen atom, it emits photon of:
Correct answer: B. 10.20 eV
- A. 3.04 eV
- B. 10.20 eV
- C. 13.6 eV
- D. 3.8 eV
Explanation
When an electron in a hydrogen atom transitions from the second orbit (n=2) to the first orbit (n=1), it emits a photon with a specific energy. This energy corresponds to the difference in energy levels between these two orbits. According to the Bohr model, the energy ( E ) of an electron in a particular orbit ( n ) is given by: E(n)=−13.6 eV /n2 The energy of the photon emitted during the transition is the difference in energy between the two levels: Ephoton =E1−E2 = -Eo/n2 =−13.6 eV(1/12 −1/22 ) Ephoton =13.6 eV(1/4 - 1 ) =13.6 eV×3/4 =10.20 eVSo, the photon emitted has an energy of 10.20 eV. This is the characteristic energy of the photon emitted in the Lyman series of the hydrogen spectrum when an electron falls from the second to the first orbit.
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