Question

A narrow beam of singly-charged carbon ions, moving at a constant velocity of 6.0 × 10⁴ m s⁻¹, is sent perpendicularly in a rectangular region of uniform magnetic field B = 0.5 T (figure). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1.6 × 10⁻²⁷) kg, where A is the mass number.

Answer 1

Given:
Velocity of a narrow beam of singly-charged carbon ions, v = 6.0 × 10⁴ m s⁻¹
Strength of magnetic field B = 0.5 T
Separations between the two beams from the incident beam are 3.0 cm and 3.5 cm.
Mass of an ion = A(1.6 × 10⁻²⁷) kg


The radius of the curved path taken by the first beam, $r_1=\frac{3}{2}=1.5\mathrm{cm}$
The radius of the curved path taken by the second beam, $r_2=\frac{3.5}{2}\mathrm{cm}$
We know:
$r_1=\frac{m_{1}v}{qB}$,
where m₁ is the mass of the first isotope and q is the charge.

For the second beam:
$r_2=\frac{m_{2}v}{qB}$,
where m₂ is the mass of the first isotope and q is the charge.

$$\begin{gathered} \frac{r_1}{r_2}=\frac{m_1}{m_2} \\ \frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{A_1\times 1.6\times {10}^{-27}}{A_2\times 1.6\times {10}^{-27}} \\ \frac{6}{7}=\frac{A_1}{A_2} \end{gathered}$$
As
$r_1=\frac{m_{1}v}{qB}$
$$\begin{gathered} \Rightarrow m_1=\frac{qB{r}_1}{v} \\ =\frac{1.6\times {10}^{-19}\times 0.5\times 0.015}{6\times {10}^4} \\ =20\times {10}^{-27}\mathrm{kg} \\ =\frac{20\times {10}^{-27}}{1.6\times {10}^{-27}}\mathrm{u} \\ =12.5\mathrm{u} \end{gathered}$$
Also,
$$\begin{gathered} A_2=\frac{7}{6}{A}_1 \\ =\frac{7}{6}\times 12.5 \\ =14.58\mathrm{u} \end{gathered}$$
So, the two isotopes of carbon used are ¹²C₆ and ¹⁴C₆.