English

State Bohr’S Postulate of Hydrogen Atom Which Successfully Explains the Emission Lines in the Spectrum of Hydrogen Atom - Physics

Advertisements
Advertisements

Question

State Bohr’s postulate of hydrogen atom which successfully explains the emission lines in the spectrum of hydrogen atom. Use Rydberg formula to determine the wavelength of Hα line. [Given: Rydberg constant R = 1.03 × 107 m−1]

Solution

Bohr’s third postulate successfully explains emission lines. It states that ‘Whenever an electron jumps from one of its specified non-radiating orbit to another such orbit, it emits or absorbs a photon whose energy is equal to the energy difference between the initial and final states’.

Thus,

`E_1-E_t=hv=(hc)/lambda`

The Rydberg formula for the Balmer series is given as

 `1/lambda=R(1/n_t^2-1/n_1^2)`

‘R’ is a constant called the Rydberg constant and its value is 1.03 × 107 m−1.

The Hα-line of the Balmer series is obtained when an electron jumps to the second orbit (nf = 2) from the third orbit (ni = 3).

`1/lambda=1.03xx10^7(1/2^2-1/3^3)`

`1/lambda=1.03xx10^7(1/4-1/9)`

`1/lambda=1.03xx10^7((9-4)/(9xx4))`

`1/lambda=1.03xx10^7(5/36)`

`lambda=36/(5xx1.03xx10^7)`

λ = 6.99 x 10-7 

λ = 699 x 10-2 x 10-7 

λ = 699 nm

The value of λ lies in the visible region.

shaalaa.com
  Is there an error in this question or solution?
2014-2015 (March) Panchkula Set 3

RELATED QUESTIONS

The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?


Radiation from hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. Work function of cesium is 1.9 eV.


The spectral line obtained when an electron jumps from n = 5 to n = 2 level in hydrogen atom belongs to the ____________ series.


A particle has a mass of 0.002 kg and uncertainty in its velocity is 9.2 × 10−6 m/s, then uncertainty in position is ≥ ____________.

(h = 6.6 × 10−34 J s)


The energy of an electron in an excited hydrogen atom is - 3.4 eV. Calculate the angular momentum of the electron according to Bohr's theory. (h = 6.626 × 10-34 Js)


Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but same orbital angular momentum according to the Bohr model? Justify your answer.


The ratio of the ionization energy of H and Be+3 is ______.


Given below are two statements:

Statements I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.

Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increase with a decrease in principal quantum number.
In light of the above statements, choose the most appropriate answer from the options given below:


The first ionization energy of H is 21.79 × 10-19 J. The second ionization energy of He atom is ______ × 10-19J.


A 20% efficient bulb emits light of wavelength 4000 Å. If the power of the bulb is 1 W, the number of photons emitted per second is ______.

[Take, h = 6.6 × 10-34 J-s]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×