Question

Calculate the radius of second Bohr orbit in hydrogen atom from the given data.
Mass of electron= $9.1\times {10}^{-31}Kg$
Charge on the electron$=1.6\times {10}^{-19}C$
Planck's constant=$6.63\times {10}^{-34}J-s$
Permitivitty of free space = $8.85\times {10}^{-12}{C}^2/N{m}^2$

Answer 1

$r=\frac{n^2{h}^2{\epsilon }_o}{\pi mZ{e}^2}$
for hydrogen $Z=1$
$\therefore$ $r=\frac{n^2{h}^2{\epsilon }_o}{\pi m{e}^2}$
$\Rightarrow \frac{2^2\times {(6.63\times {10}^{-34})}^2\times (8.85\times {10}^{-12})}{3.14\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$
$=\frac{1556.07\times {10}^{-80}}{73.15\times {10}^{-69}}=21.3\times {10}^{-11}m$