Question

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

Answer 1

Answer:

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