The energy of a photon emitted during a transition in the Bohr hydrogen atom is given by the equation:
E = -13.6 eV * (1/n_i^2 - 1/n_f^2)
Where n_i is the initial energy level and n_f is the final energy level. The higher the energy level, the lower the energy of the photon emitted.
In this case, we need to find the transition that results in the emission of the lowest-energy photon. Using the given options, we can calculate the energies of the photons emitted:
a) n = 1 to n = 6:
E = -13.6 eV * (1/1^2 - 1/6^2) = -13.6 eV * (1 - 1/36) = -13.2 eV
b) n = 5 to n = 1:
E = -13.6 eV * (1/5^2 - 1/1^2) = -13.6 eV * (1/25 - 1) = 10.88 eV
c) n = 6 to n = 1:
E = -13.6 eV * (1/6^2 - 1/1^2) = -13.6 eV * (1/36 - 1) = 12.067 eV
d) n = 3 to n = 5:
E = -13.6 eV * (1/3^2 - 1/5^2) = -13.6 eV * (1/9 - 1/25) = -2.533 eV
e) n = 1 to n = 5:
E = -13.6 eV * (1/1^2 - 1/5^2) = -13.6 eV * (1 - 1/25) = -12.944 eV
Based on these calculations, option b) (n = 5 to n = 1) results in the emission of the lowest-energy photon (-10.88 eV).
Of the following transitions in the Bohr hydrogen atom, which of the following results in the emission of the lowest-energy photon.
Question 2 options:
a) n = 1 to n = 6
b) n = 5 to n = 1
c) n = 6 to n = 1
d) n = 3 to n = 5
e) n = 1 to n = 5
1 answer