Question

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

Answer 1

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:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = 2 in an hydrogen atom:

$E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV$

Energy of n = 3 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy of n = 4 in an hydrogen atom:

$E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV$

Energy of n = 5 in an hydrogen atom:

$E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV$

a) n = 2 to n = 4 (absorption)

$\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV$

b) n = 2 to n = 1 (emission)

$\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV$

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

$\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV$

d) n = 4 to n = 3 (emission)

$\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV$

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

$E=h\nu$

h = Planck's constant

$\nu$ frequency of the wave

So, the increasing order of magnitude of the energy difference :

$E_4$

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