Final answer:
The energy of photons emitted during a hydrogen atom's transition from the 4th to the 2nd energy state is 2.55 eV.
Step-by-step explanation:
To calculate the energy of photons emitted by a hydrogen atom transitioning from the 4th to the 2nd energy state, we use the Rydberg formula for the energy of a photon emitted or absorbed:
Energy = -13.6 eV (1/n_f^2 - 1/n_i^2)
where:
- n_f = final energy level (2 for the 2nd energy state)
- n_i = initial energy level (4 for the 4th energy state)
- 13.6 eV is the energy of an electron in the ground state (n=1) of the hydrogen atom
Plugging the numbers into the formula gives:
Energy = -13.6 eV (1/2^2 - 1/4^2)
Energy = -13.6 eV (1/4 - 1/16)
Energy = -13.6 eV (4/16 - 1/16)
Energy = -13.6 eV (3/16)
Energy = -13.6 eV * 0.1875
Energy = -2.55 eV
The negative sign indicates emission of energy, so the energy of the emitted photon is 2.55 eV.