In a nuclear experiment a proton with kinetic energy 3.0 MeV moves in a circular path in a uniformmagnetic field. What energy must the following two particles haveif they are to circulate in the same orbit?

(a) an alpha particle (q = + 2e, m = 4.0 u)

1 MeV

(b) a deuteron (q = + e, m = 2.0 u)

2 MeV

a) K = 3 MeV b) K = 1.5 MeV

Explanation:

We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.

F = q v x B

We can also write this equation based on the modules of the vectors

F = qv B sin θ

With Newton's second law

F = ma

F = m v² / r

q v B = m v² / r

v = q B r / m

The kinetic energy is

K = ½ m v²

Substituting

K = ½ m (q B r / m) ²

K = ½ B² r² q² / m

K = (½ B² R²) q²/m

The amount in brackets does not change during the experiment

K = A q² / m

For the proton

K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J

With this data we can find the amount we call A

A = K m/q²

A = 4.8 10⁻¹³ 1.67 10⁻²⁷ / (1.6 10⁻¹⁹) ²

A = 3.13 10⁻²

With this value we can write the equation

K = 3.13 10⁻² q² / m

Alpha particle

m = 4 uma = 4 1.66 10⁻²⁷ kg

K = 3.13 10⁻² (2 1.6 10⁻¹⁹) ² / 4.0 1.66 10⁻²⁷

K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV

K = 3 MeV

Deuteron

K = 3.13 10⁻² (1.6 10⁻¹⁹) ²/2 1.66 10⁻²⁷

K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)

K = 1.5 10⁶ eV

K = 1.5 MeV