![Q.60 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 in the magnetic field. The radius of its path in the magnetic field (correct to two decimal places) is _______ cm. [mass of deuteron is 3.34 × 10⁻²⁷ kg, 1 MeV = 1.6 × 10⁻¹³ J and e = 1.6 × 10⁻¹⁹ C]](https://www.letstalkacademy.com/wp-content/uploads/2025/12/slide_60-12.jpg)
Q.60 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 in the magnetic field. The radius of its path in the magnetic field (correct to two decimal places) is _______ cm.
[mass of deuteron is 3.34 × 10⁻²⁷ kg, 1 MeV = 1.6 × 10⁻¹³ J and e = 1.6 × 10⁻¹⁹ C]
Key Formula
$$ r = \frac{mv}{qB} $$
Step-by-Step Solution
- Convert kinetic energy to joules:
\( KE = 17 \times 1.6 \times 10^{-13} = 2.72 \times 10^{-12} \) J - Calculate velocity from \( KE = \frac{1}{2}mv^2 \):
\( v = \sqrt{\frac{2 \times 2.72 \times 10^{-12}}{3.34 \times 10^{-27}}} = 4.04 \times 10^7 \) m/s - Substitute into radius formula with deuteron charge \( q = e = 1.6 \times 10^{-19} \) C:
\( r = \frac{3.34 \times 10^{-27} \times 4.04 \times 10^7}{1.6 \times 10^{-19} \times 2.4} = 0.351 \) m = 35.10 cm
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Introduction
Discover how to precisely calculate the deuteron radius in mass spectrometer with 17 MeV kinetic energy entering a 2.4 T magnetic field. This deuteron path radius problem, using mass 3.34×10⁻²⁷ kg and e=1.6×10⁻¹⁹ C, is ideal for CSIR NET Life Sciences and physics aspirants mastering charged particle motion.
Physics Behind Deuteron Motion
Deuterons (²H nuclei) in mass spectrometers follow circular paths when velocity is perpendicular to the magnetic field. The radius depends on mass, velocity from kinetic energy, charge, and field strength.
Key equation: \( r = \frac{mv}{qB} \), where velocity \( v = \sqrt{\frac{2KE}{m}} \).
Common Mistakes to Avoid
- Forgetting to convert MeV to joules using \( 1.6 \times 10^{-13} \) J/MeV.
- Using proton charge instead of deuteron (still +e).
- Incorrect units: Ensure radius in cm to two decimals.
CSIR NET Exam Tips
Practice similar charged particle in magnetic field problems. Relate to ion trajectories in cyclotrons or velocity selectors. Verify non-relativistic assumption: \( v \approx 0.135c \), valid here.
