Find the energy of the photon released in the transition from n₁ = 6 to n₂ = 1 for a hydrogen atom. (Note: Use Rydberg Formula)
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- When an electron moves from a higher state to a lower state it will release energy in the form of electromagnetic radiation. The wavelength of the released photon is given by the Rydberg formula.
- Rydberg formula is given by,
Here R∞ is the Rydberg constant and has a value of 1.097373 × 107 m-1, and n1 and n2 are the states in which transition takes place.
- Combining this equation with the energy of the photon we get,
Here n1 and n2 are the states in which transition takes place.
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- Student A & B are studying the Zeeman effect. They observe that the energy of an electron in the p-level of an atom is changed in the presence of a magnetic field of magnitude 4.6 T. What is the difference between the largest and smallest possible energies? (bohr magneton = μB = 9.27x10-24 J/T).Why does the transition of Ninitial=4 to Nfinal=5 of a hydrogen atom result in the absorption of the longest wavelength photon?Chapter 39, Problem 044 A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of -1.51 eV makes a transition to a state with an excitation energy (the difference between the energy of the state and that of the ground state) of 10.200 eV. (a) What is the energy of the photon emitted as a result of the transition? What are the (b) higher quantum number and (c) lower quantum number of the transition producing this emission? Use -13.60 eV as the binding energy of an electron in the ground state. (a) Number Units (b) Number Units (c) Number Units
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