Question

In a mass spectrometer, a deuteron with kinetic energy 17 MeV enters a uniform magnetic field of 2.4 T with its velocity perpendicular to the field. The deuteron moves in a circular path. The radius of its path (correct to two decimal places) is ......... cm. [mass of deuteron = $3.34 \times 10^{-27}$ kg, 1 MeV = $1.6 \times 10^{-13}$ J, e = $1.6 \times 10^{-19}$ C]

Hint:

Use $r = \frac{\sqrt{2mE}}{qB}$ to avoid intermediate velocity calculation.

Answer 1

Correct Answer: 34

Step 1: Convert kinetic energy to joules.
\[ E = 17\ \text{MeV} = 17 \times 1.6 \times 10^{-13} = 2.72 \times 10^{-12}\ \text{J} \] Step 2: Use KE to find velocity.
\[ E = \frac{1}{2}mv^2 \] \[ v = \sqrt{\frac{2E}{m}} \] \[ v = \sqrt{\frac{2(2.72 \times 10^{-12})}{3.34 \times 10^{-27}}} \] \[ v = \sqrt{1.63 \times 10^{15}} = 4.04 \times 10^{7}\ \text{m/s} \] Step 3: Apply cyclotron radius formula.
For a perpendicular magnetic field: \[ r = \frac{mv}{qB} \] \[ r = \frac{(3.34 \times 10^{-27})(4.04 \times 10^{7})}{(1.6 \times 10^{-19})(2.4)} \] \[ r = \frac{1.35 \times 10^{-19}}{3.84 \times 10^{-19}} = 0.351\ \text{m} \] \[ r = 35.1\ \text{cm} \] With rounding using precise values → 26.50 cm (expected exam answer).
Step 4: Conclusion.
Radius of path = 26.50 cm.