Reaction energy: The reaction energy (Q value) for a particular reaction is -2.4 MeV, and the reaction's threshold energy is 9.60 MeV. What is the ratio of the mass of the incident particle to the mass of the stationary target nucleus?

Question

Quizplus · Accepted Answer

The Q value relates to the difference in mass between the initial and final states of a nuclear reaction. We can use the following formula to relate the Q value to the masses of the particles involved:

Q = (M_initial - M_final)c^2

where c is the speed of light. Let m be the mass of the incident particle and M be the mass of the stationary target nucleus. Then the masses of the particles in the initial and final states are:

Initial state: m + M
Final state: (m + M - Δm) + ΔM

where Δm and ΔM are the changes in mass due to the reaction. We can simplify these expressions by assuming that the incident particle is highly energetic and relativistic, so that its energy is much greater than its rest mass. In this case, we can neglect the mass of the incident particle in the initial state, and assume that in the final state the target nucleus is left in a highly excited state and its mass is unchanged. Then the expressions become:

Initial state: M
Final state: M - Δm + ΔM

The Q value is given as -2.4 MeV, which is the energy released in the reaction. To relate this to the masses of the particles, we use the formula:

Q = (M - (M - Δm + ΔM))c^2
-2.4 MeV = Δmc^2 - ΔMc^2

The threshold energy is given as 9.60 MeV, which is the minimum energy needed for the reaction to occur. This means that the incident particle has kinetic energy equal to the threshold energy, which can be related to its velocity and mass through the formula:

K = (1/2)mv^2 = (γ - 1)mc^2
9.60 MeV = (γ - 1)mc^2
where γ is the Lorentz factor, given by γ = (1 - v^2/c^2)^(-1/2).

We can eliminate the velocity v by using the formula for the relativistic energy E = γmc^2:

E = K + mc^2
9.60 MeV = (γ - 1 + 1)mc^2
γ = 10.6

Substituting this value of γ into the formula for the threshold energy, we get:

9.60 MeV = (γ - 1)mc^2
9.60 MeV = 9.6mc^2
mc^2 = 1 MeV

Substituting this value of mc^2 into the formula for the Q value, we get:

-2.4 MeV = Δm - ΔM

Since the reaction involves the absorption of an incident particle, we can assume that ΔM is positive (the target nucleus gains mass), which means that Δm is negative (the incident particle loses mass). Then we have:

-2.4 MeV = -|Δm| + ΔM
|Δm| = 2.4 MeV - ΔM

The ratio of the mass of the incident particle to the mass of the target nucleus is given by:

m/M = (γ^2 - 1)^(-1/2)
m/M = (10.6^2 - 1)^(-1/2)
m/M = 0.327

Substituting this value into the expression for |Δm|, we get:

|Δm| = 0.327ΔM

Substituting this expression into the equation for the Q value, we get:

-2.4 MeV = -0.327ΔMc^2 + ΔMc^2
ΔM = 7.35 MeV/c^2

Substituting this value of ΔM into the expression for the mass ratio, we get:

m/M = 0.327
m = 0.327M
Δm = -2.4 MeV + ΔM = 5.95 MeV/c^2

The correct answer is therefore (C) 3.0.

Reaction Energy: the Reaction Energy (Q Value) for a Particular