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:

2.55 eV is absorbed.

Explanation:

The energy of the nth state of hydrogen atom is,

$E_{n}=-\frac{13.6}{n^{2} }$

Now energy for the second state is,

$E_{2}=-\frac{13.6}{2^{2} }\\E_{2}=-3.4eV$

Now energy for the fourth state is,

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

Now energy required to go from 2nd state to 4th state is,

$hf+E_{2}=E_{4}\\ hf=E_{4}-E_{2}$

Therefore,

$hf=-0.85eV-(-3.4eV)\\hf=2.55eV$

Therefore, 2.55 eV energy is absorbed by the hydrogen atom to from n=2 state to n=4 state.