Using Bohr’s postulates, derive the expression for the radius of the nth orbit in which the electron is revolving in hydrogen atom. How does  de-Broglie’s hypothesis explain the stability of hydrogen atom? Explain. 

 

 
 
 
 
 

Answers (1)
S safeer

According to Bohr's postulates, 

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

L_n\rightarrow  Angular momentum

v_n\rightarrow Speed of moving electron in n^{th}} orbit

r_n\rightarrow radius of n^{th} orbit

h\rightarrow Planck's constant

m\rightarrow mass of particle

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}

According to de Broglie's Hypothesis, material particles such as electrons also have wave nature. 

For an electron moving in n^{th} circular orbit of radius r_n, the total distance is the circumference of the orbit.

Thus, 

n\lambda =2\pi r\: ;\: n=1,2,3...

\lambda \rightarrow de - Broglie wavelength of an electron moving in n^{th} orbit.

\lambda =\frac{h}{p}

p\rightarrow  the magnitude of the electron's momentum

h\rightarrow Planck's constant

If the speed of the electron is much less than the speed of light then,

p=mv_n

\lambda =\frac{h}{mv_n}

\frac{nh}{mv_n}=2\pi r_n

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

This is the quantum condition proposed by Bohr for the angular momentum of the electron and this explains the stability of hydrogen atom.

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