Question

Prove that the radius of nth bohr's orbit of an atom is directly proportional to n2

Answer 1

Explanation:

here the formula for radius is directly proportional to n^2

Answer 2

Yes, radius of nth orbit of hydrogen atom is directly proportional to n^2 as :---

As per Coulomb's law:

F = $\frac{q_{1} q_{2}}{4\pi \epsilon r^{2}}$

$\frac{1}{4\pi \epsilon }$ = 9 x $10^{9}$ (in MKS)

     = 1          (in CGS)

Force of attraction betwween electron and nucleus

F = $\frac{K z e^{2} }{r^{2} }$

Centrifugal force balances electrostatic force

$\frac{mv^{2} }{r}$ =  $\frac{K z e^{2} }{r^{2} }$          

m$v^{2}$ = $\frac{Kze^{2} }{r}$               ....(I)

But according to Bohr

mvr = $\frac{nh}{2\pi }$

v = $\frac{nh}{2\pi r}$

$v^{2}$ = $\frac{n^{2}h^{2} }{4\pi ^{2} m^{2} r^{2} }$

Putting value in (i)

$\frac{m n^{2}h^{2} }{4\pi ^{2} m^{2} r^{2} }$ = $\frac{Kze^{2} }{r}$

r = $\frac{n^{2}h^{2} }{4\pi ^{2} m Kze^{2} }$

r = $\frac{0.529 n^{2} }{Z}$ angstrom

Radius of first shell of Hydrogen is 0.529 angstrom