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Obtain Expressions for Longest and Shortest Wavelength of Spectral Lines in Ultraviolet Region for Hydrogen Atom - HSC Science (Computer Science) 12th Board Exam - Physics

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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, En = Energy of electron in nth higher orbit

         Ep = Energy of electron in pth lower orbit

ii. According to Bohr’s third postulate,

En - Ep = hv

∴ v = (Ea - Ep)/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`

 

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APPEARS IN

 2014-2015 (October) (with solutions)
Question 5.2 | 7.00 marks
Solution Obtain Expressions for Longest and Shortest Wavelength of Spectral Lines in Ultraviolet Region for Hydrogen Atom Concept: Bohr'S Model for Hydrogen Atom.
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