Question

Fe+ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A (1.6 × 10−27) kg, where A is the mass number.

Answer 1

The circular path’s radius is 120 cm.

Explanation:

Potential difference by which there is an acceleration of the Fe+ ions, V = 500 V

Homogeneous magnetic field’s strength, B = 20 × $10^{-3}$ T

Two isotopes mass numbers are m$_1$=57 and m$_2$=58

Ion’s mass = A (1.6 × $10^{-27}$) kg

Step 1:

It is understood that the circular path’s radius explained in a magnetic field by a particle,

$r_{1}=\frac{m_{1} v_{1}}{q B}$

$\begin{aligned}&r_{2}=\frac{m_{2} v_{2}}{q B}\\&\frac{r_{1}}{r_{2}}=\frac{m_{1} v_{1}}{m_{2} v_{2}}\end{aligned}$

The acceleration through the same potential V as both the isotopes, the kinetic energy increased by the two particles will be identical.

Step 2:

Force developed by potential difference is given by

$\boldsymbol{F}=\boldsymbol{q} \boldsymbol{V}$

where v = applied potential

q = charge

$\begin{aligned}&\Rightarrow q V=\frac{1}{2} m_{1} v_{1}^{2}=\frac{1}{2} m_{2} v_{2}^{2}....equation 1\\&=\frac{v_{1}^{2}}{v_{2}^{2}}\\&\Rightarrow \frac{r_{1}}{r_{2}}=\left(\frac{m_{1}}{m_{2}}\right)^{\frac{3}{2}} equation....2\end{aligned}$

Also we have

$r_{1}=\frac{m_{1} v_{1}}{q B}$

Step 3:

From equation ...1 we get

$\begin{aligned}&r_{1}=\frac{m_{1} \sqrt{2 q V m_{1}}}{q B}\\&=\frac{1}{B} \frac{\sqrt{2 V m_{1}}}{q}\\&=\frac{\sqrt{1000 \times 57 \times 1.6 \times 10^{-27}}}{\sqrt{1.6 \times 10^{-19} \times 20 \times 10^{-3}}}\\&=1.19 \times 10^{-2} \mathrm{m}\\&=119 \mathrm{cm}\end{aligned}$

Step 4:

By calculating the radius of the isotope 2

From equation.....2 we get

$\begin{aligned}&\frac{r_{1}}{\left(\frac{m_{1}}{m_{2}}\right)^{\frac{3}{2}}}=r_{2}\\&=\left(\frac{58}{57}\right)^{\frac{3}{2}} \times 119 \mathrm{cm}\\&=120 \mathrm{cm}\end{aligned}$