Respuesta :
Answer:
The n = 2 → n = 3 transition results in the absorption of the highest-energy photon.
Explanation:
[tex]E_n=-13.6\times \frac{Z^2}{n^2}ev[/tex]
Formula used for the radius of the [tex]n^{th}[/tex] orbit will be,
where,
[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit
n = number of orbit
Z = atomic number
Here: Z = 1 (hydrogen atom)
Energy of the first orbit in H atom .
[tex]E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV[/tex]
Energy of the second orbit in H atom .
[tex]E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV[/tex]
Energy of the third orbit in H atom .
[tex]E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV[/tex]
Energy of the fifth orbit in H atom .
[tex]E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV[/tex]
Energy of the sixth orbit in H atom .
[tex]E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV[/tex]
Energy of the seventh orbit in H atom .
[tex]E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV[/tex]
During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :
1) n = 2 → n = 3
2) n= 5 → n = 6
Energy absorbed when: n = 2 → n = 3
[tex]E=E_3-E_2[/tex]
[tex]E=(-1.51 eV) -(-3.40 eV)=1.89 eV[/tex]
Energy absorbed when: n = 5 → n = 6
[tex]E'=E_6-E_5[/tex]
[tex]E'=(-0.378 eV)-(-0.544 eV) =0.166 eV[/tex]
1.89 eV > 0.166 eV
E> E'
So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.
