Respuesta :
The question is incomplete , complete question is:
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 4 to n = 3
Answer:
Hence the order of the transition will be : d < a < c < b
Explanation:
[tex]E_n=-13.6\times \frac{Z^2}{n^2}ev[/tex]
where,
[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
[tex]E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV[/tex]
Energy of n = 2 in an hydrogen atom:
[tex]E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV[/tex]
Energy of n = 3 in an hydrogen atom:
[tex]E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV[/tex]
Energy of n = 4 in an hydrogen atom:
[tex]E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV[/tex]
Energy of n = 5 in an hydrogen atom:
[tex]E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV[/tex]
a) n = 2 to n = 4 (absorption)
[tex]\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV[/tex]
b) n = 2 to n = 1 (emission)
[tex]\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV[/tex]
Negative sign indicates that emission will take place.
c) n = 2 to n = 5 (absorption)
[tex]\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV[/tex]
d) n = 4 to n = 3 (emission)
[tex]\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV[/tex]
Negative sign indicates that emission will take place.
According to Planck's equation, higher the frequency of the wave higher will be the energy:
[tex]E=h\nu [/tex]
h = Planck's constant
[tex]\nu [/tex] frequency of the wave
So, the increasing order of magnitude of the energy difference :
[tex]E_4<E_1<E_3<E_2[/tex]
And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :
: d < a < c < b
