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
