# 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)

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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