Question

In Bohr's Model of hydrogen atom, the electron circulates around the nucleus in a path of radius $5.1\times {10}^{-11}\text{m}$ at a frequency of $6.8\times {10}^{15}\text{Hz}$. The value of magnetic field and equivalent magnetic moment will be

• A. $15.4\text{T}$ and $8.9\times {10}^{-24}{\text{Am}}^2$
• B. $18.4\text{T}$ and $9.8\times {10}^{-24}{\text{Am}}^2$
• C. $13.4\text{T}$ and $8.9\times {10}^{-24}{\text{Am}}^2$
• D. $18.4\text{T}$ and $8.9\times {10}^{-24}{\text{Am}}^2$

Answer 1

The correct option is C $13.4\text{T}$ and $8.9\times {10}^{-24}{\text{Am}}^2$

Magnetic field at the centre,

$B=\frac{{\mu }_{0}i}{2R}$

$\because i=q\nu$

So, $B=\frac{{\mu }_{0}q\nu }{2R}$

$\Rightarrow B=\frac{4\pi \times {10}^{-7}\times 1.6\times {10}^{-19}\times 6.8\times {10}^{15}}{2\times 5.1\times {10}^{-11}}$

$\Rightarrow B=13.4\text{T}$

The current carrying circular loop is equivalent to a magnetic dipole with magnetic moment,

$\Rightarrow M=NiA$

$\Rightarrow M=N\left(q\nu \right)\left(\pi R^2\right)$

$\Rightarrow M=\left(1\right)(1.6\times {10}^{-19}\times 6.8\times {10}^{15})[3.14\times (5.1\times {10}^{-11}{)}^2]$

$\therefore M=8.9\times {10}^{-24}{\text{A-m}}^2$

Hence, $\left(\text{C}\right)$ is the correct answer.