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  • (a) State the postulates of Bohr’s model of hydrogen atom and derive the expression for Bohr radius. (b) Find the ratio of the longest and the shortest wavelengths amongst the spectral lines of Balmer series in the spectrum of hydrogen

(a) State the postulates of Bohr’s model of hydrogen atom and derive the expression for Bohr radius.

(b) Find the ratio of the longest and the shortest wavelengths amongst the spectral lines of Balmer series in the spectrum of hydrogen atom.

 

 
 
 
 
 

Answers (1)

a) 

Postulate I
Bohr postulated that in an atom, electron/s could revolve in stable orbits without emitting radiant energy and each atom can exist in certain stable states. Also, each state has a definite total energy. These are stationary states of the atom.

Postulate II
Bohr defined these stable orbits in his second postulate. According to this postulate:

An electron revolves around the nucleus in orbits
The angular momentum of revolution is an integral multiple of h/2pi – where h is Planck’s constant [h = 6.6 x 10-34 Js].
Hence, the angular momentum (L) of the orbiting electron is: L = nh/2pi

Postulate III
According to this postulate, an electron can transition from a non-radiating orbit to another of a lower energy level. In doing so, a photon is emitted whose energy is equal to the energy difference between the two states.

 

According to Bohr's postulates, 

L_n=mv_nr_n=\frac{nh}{2\pi }

For a dynamically stable orbit in a hydrogen atom, 

F_e=F_c

 \frac{mv_n^2}{r_n}=\frac{1}{4\pi \epsilon _0}\frac{e^2}{r_n^2}

mv_n^2=\frac{1}{4\pi \epsilon _0}\frac{e^2}{r_n}

v_n^2=\frac{e^2}{4\pi \epsilon _0mr_n}

v_n=\frac{e}{\sqrt{4\pi \epsilon _0mr_n}}\: \: \rightarrow (1)

Also, v_n=\frac{nh}{2\pi mr_n}\: \: \rightarrow (2)

Equating (1) and (2),

\frac{e}{\sqrt{4\pi \epsilon _0 mr_n}}=\frac{nh}{2\pi mr_n}

\frac{e^2}{4\pi \epsilon _0 mr_n}=\frac{n^2h^2}{4\pi^2 m^2r_n^2}

\frac{e^2}{\epsilon _0}=\frac{n^2h^2}{\pi mr_n}

r_n=\frac{n^2h^2\epsilon _0}{\pi me^2}

b) 

\\\frac{1}{\lambda _{max}}=R(\frac{1}{2^2}-\frac{1}{3^2})\\\\\frac{1}{\lambda _{max}}=R\frac{5}{36}

 Now calculation for smallest wavelength, we have :

\frac{1}{\lambda _{smallest}}=R\left \{ \frac{1}{2^{2}}- \frac{1}{\infty ^{2}}\right \}

\frac{1}{\lambda _{smallest}}=\frac{R}{4}

\frac{\lambda_{max}}{\lambda_{smallest}}=\frac{9}{5}

 

Posted by

Safeer PP

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