What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (Rydberg constant R = 1.097 x 10^7 m^-1) (2021)
Practice Questions
1 question
Q1
What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (Rydberg constant R = 1.097 x 10^7 m^-1) (2021)
1.89 eV
3.40 eV
4.86 eV
5.13 eV
Energy = 13.6 eV * (1/n1^2 - 1/n2^2) = 13.6 * (1/2^2 - 1/3^2) = 3.40 eV.
Questions & Step-by-step Solutions
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Q
Q: What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (Rydberg constant R = 1.097 x 10^7 m^-1) (2021)
Solution: Energy = 13.6 eV * (1/n1^2 - 1/n2^2) = 13.6 * (1/2^2 - 1/3^2) = 3.40 eV.
