Question

A narrow beam of singly-charged carbon ions, moving at a constant velocity of $6.0\times {10}^{4}m{s}^{-1,}$ is sent perpendicularly in a rectangular region having uniform magnetic field B = 0.5 T (figure 34-E15). 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 = $A(1.6\times {10}^{-27})kg,$ where A is the mass number.

Answer 1

$u=6\times {10}^{4}m/s,B=0.5T,$

$\Rightarrow r_1=\frac{3}{2}=1.5cm,r_2=\frac{3.5}{2}cm$

$\Rightarrow r_1=\frac{mv}{qB}$

$\Rightarrow 1.5\times {10}^{-2}=\frac{m_1\times 6\times {10}^4}{6\times 1.6\times {10}^{-19}\times 0.5}$

$\Rightarrow m_1=1.2\times {10}^{-25}$

$=12\times {10}^{-26}$

Hence carbon is $C^{12}$

Similarly $m_2=14\times {10}^{-26}$ $r_2=A$

Taking common ratio = 2 (for carbon)

The isotopes used are $12C_6$ and $14C_6$