#### Question

State Bohr’s third postulate for hydrogen (H2) atom. Derive Bohr’s formula for the wave number. Obtain expressions for longest and shortest wavelength of spectral lines in ultraviolet region for hydrogen atom

#### Solution

Third postulate (Transition and frequency condition):-

As long as electron remains in one of the stationary orbits, it does not radiate energy. Whenever an electron jumps from higher stationary orbit to lower stationary orbit, it radiates energy equal to the difference in energies of the electron in the two orbits.

Bohr’s formula for spectral lines in hydrogen spectrum:-

i. Let, E_{n} = Energy of electron in nth higher orbit

E_{p} = Energy of electron in pth lower orbit

ii. According to Bohr’s third postulate,

E_{n} - E_{p} = hv

∴ v = (E_{a} - E_{p})/h ...........(1)

iii. `"But "E_n=-(me^4)/(8epsilon_0^2"h"^2"n"^2)` ............(2)

`E_p=-(me^4)/(8epsilon_0^2"h"^2"p"^2` .............(3)

iv. From equations (1), (2) and (3),

`"v"=(((-me^4)/(8epsilon_0^2"h"^2"n"^2))-(-(me^4)/(8epsilon_0^2"h"^2"p"^2)))/h`

`therefore"v"=(me^4)/(8epsilon_0^2h^3)[-1/n^2+1/p^2]`

`thereforec/lambda=(me^4)/(8epsilon_0^2h^3)[1/p^2-1/n^2]` [∵ v = c/λ]

where, c = speed of electromagnetic radiation

`therefore1/lambda=(me^4)/(8epsilon_0^2h^3c)[1/p^2-1/n^2]`

v. `1/lambda=R[1/p^2-1/n^2]` ............(4)

`"where, "(me^4)/(8epsilon_0^2h^3c)=R="Rydberg’s constant"`

Equation (4) represents Bohr’s formula for hydrogen spectrum.

vi. 1/λ is called wave number (`bar"v"` ) of the line.

`thereforebar"v"=1/lambda=R(1/p^2-1/n^2)`

For longest wavelength in ultraviolet region (Lyman series),

p = 1; n = 2

`therefore1/lambda_(L_1)=R(1/1^2-1/2^2)=(3R)/4`

`thereforelambda_(L_1)=4/(3R)`

For shortest wavelength in ultraviolet region

P = 1; n = ∞

`therefore1/lambda_(L_2)=R(1/1^2-1/oo^2)=R`

`thereforelambda_(L_2)=1/R`