Question

The energy of the ground state in the Bohr model of the hydrogen atom is -13.6 eV. In a transition from the n = 2 state to the n = 4 state, a photon of energy: (A) 3.40 eV is emitted (B) 3.40 eV is absorbed (C) 2.55 eV is emitted (D) 2.55 eV is absorbed

Answer 1

Answer:

(D) 2.55 eV is absorbed

Explanation:

For an electron located in an inner orbit to jump an upper orbit, it needs an energy equal to the difference between the two quantized orbits. This energy comes from the absorption of a photon with the same energy required. The formula for the energy levels of a Hydrogen atom is:

$E{n}=\frac{E_{0}}{n^2}$

where $E_{0}$ is the energy of the ground state and n is the energy level or state. Finding $E_{4}$ and $E_{2}$:

$E{4}=\frac{-13.6 eV}{4^2}=-0.85 eV\\E{2}=\frac{-13.6 eV}{2^2}=-3.40 eV$

Finally we find the energy difference between n=4 and n=2 state:

$E{4}-E{2}=-0.85 eV -(-3.40 eV)=2.55 eV}$