Question

How many times does the electron of hydrogen atom go round the first Bohr orbit in one $\text{second}$?
$[m=9.1\times {10}^{-31}\text{kg},e=1.6\times {10}^{-19}\text{C}{\epsilon }_0=8.85\times {10}^{-12}{\text{F m}}^{-1},h=6.6\times {10}^{-34}\text{J s}]$

• A. $6.6\times {10}^5$
• B. $6.6\times {10}^{10}$
• C. $6.6\times {10}^{15}$
• D. $6.6\times {10}^{13}$

Answer 1

The correct option is C $6.6\times {10}^{15}$

Here, we have to find frequency which is given by,

$\nu =\frac{m{Z}^2{e}^4}{4{\epsilon }_0^2{n}^3{h}^3}$

For Hydrogen atom $Z=1$ and for first Bohr orbit, $n=1$

$\nu =\frac{9.1\times {10}^{-31}\times (1.6\times {10}^{-19}{)}^4}{9\times (8.85\times {10}^{-12}{)}^2\times (1{)}^3\times (6.6\times {10}^{-34}{)}^2}$

$=\frac{59.63\times {10}^{-107}}{9.00\times {10}^{-122}}=6.6\times {10}^{15}\text{Hz}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.