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

2.55 eV is absorbed.

Step-by-step explanation:

The energy of the nth state of hydrogen atom is,

`E_(n)=-(13.6)/(n^(2) )`

Now energy for the second state is,

`E_(2)=-(13.6)/(2^(2) )\\E_(2)=-3.4eV`

Now energy for the fourth state is,

`E_(4)=-(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.