AP Physics 2 Exam Question - Nuclear Reactions

A highly unstable useonium isotope, ^238U, undergoes alpha decay to produce a stable thorium isotope. The alpha particle emitted during the decay has a kinetic energy of 5.0 MeV. Assuming the atomic mass of ^238U is 238.0508 amu and ^238U has a rest energy of 1904.9738 MeV, calculate:

a) The atomic mass of the thorium isotope produced. b) The Q-value for the alpha decay process. c) The speed of the alpha particle after the decay process.

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a) The atomic mass of the thorium isotope produced can be calculated by subtracting the mass of the alpha particle from the mass of ^238U. The mass of the alpha particle is approximately equal to 4 amu.

First, convert the atomic mass of ^238U to kg:

Given: atomic mass of ^238U = 238.0508 amu

Avogadro's number, N_A = 6.022 × 10^23 molecules/mole 1 amu = 1.660539 x 10^-27 kg

Convert amu to kg:

238.0508 amu = 238.0508 x (1.660539 x 10^-27 kg/amu) = 3.952 x 10^-25 kg

Now, subtract the mass of the alpha particle (4 amu) from the mass of ^238U:

Mass of the thorium isotope = 238.0508 amu - 4 amu = 234.0508 amu

Therefore, the atomic mass of the thorium isotope produced is approximately 234.0508 amu.

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b) The Q-value for the alpha decay process can be calculated using the equation:

Q = (mass of initial nucleus) - (mass of final nucleus) - (mass of alpha particle)

Given: mass of initial nucleus (^238U) = 238.0508 amu mass of final nucleus (thorium isotope) = 234.0508 amu mass of alpha particle = 4 amu

Substitute the values into the equation:

Q = 238.0508 amu - 234.0508 amu - 4 amu

Q = 0 amu

Therefore, the Q-value for the alpha decay process is 0 amu.

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c) The speed of the alpha particle after the decay process can be calculated using the conservation of energy and applying the equation:

$KE_{\text{before}} + E_{\text{rest}} = KE_{\text{after}} + E_{\text{rest}}$

Given: initial kinetic energy, $KE_{\text{before}} = 5.0 \, \text{MeV}$ initial rest energy, $E_{\text{rest}} = 1904.9738 \, \text{MeV}$ final rest energy, $E_{\text{rest}} = 0 \, \text{MeV}$

Since the alpha particle is much smaller than the ^238U nucleus, we can ignore its rest energy.

Substitute the given values into the equation and solve for the final kinetic energy:

$5.0 \, \text{MeV} + 1904.9738 \, \text{MeV} = KE_{\text{after}} + 0 \, \text{MeV}$

$KE_{\text{after}} = 1904.9738 \, \text{MeV} - 5.0 \, \text{MeV}$

$KE_{\text{after}} = 1899.9738 \, \text{MeV}$

Therefore, the speed of the alpha particle after the decay process can be calculated by converting the kinetic energy to speed:

$KE_{\text{after}} = \frac{1}{2} m v^2$

Convert the kinetic energy to joules: $1899.9738 \, \text{MeV} = 1899.9738 \times 1.602 \times 10^{-13} \, \text{J}$

Assuming the alpha particle has a mass of approximately 4 amu: $m = 4 \times 1.6605 \times 10^{-27} \, \text{kg}$

Substitute the values into the equation and solve for the speed, $v$:

$1899.9738 \times 1.602 \times 10^{-13} \, \text{J} = \frac{1}{2} \times 4 \times 1.6605 \times 10^{-27} \, \text{kg} \times v^2$

$v^2 = \frac{2 \times 1899.9738 \times 1.602 \times 10^{-13} \, \text{J}}{4 \times 1.6605 \times 10^{-27} \, \text{kg}}$

$v^2 = 4.581 \times 10^{-5} \, \text{m}^2/\text{s}^2$

$v \approx 0.0068 \, \text{m/s}$

Therefore, the speed of the alpha particle after the decay process is approximately 0.0068 m/s.

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Note: Calculations are done using rounded values for ease of understanding. In actual exam solutions, precise values should be used for accurate calculations.