Alpha decay ##Q##-value, recoil energy, momentum

In summary: So, in this case where we have ##\alpha##-decay and we are given atomic masses we have to calculate the nuclear masses and then calculate the Q value. In summary, when calculating the ##Q##-value for ##^{230}Th\to \; 226Ra + \alpha##, we must use nuclear masses instead of atomic masses in order to account for the fact that the nuclear charge changes in an alpha decay reaction. We must also be consistent in using either atomic or nuclear masses throughout the calculation.
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Homework Statement


Calculate the ##Q##-value for
##^{230}Th\to \; 226Ra + \alpha##.
Also calculate the kinetic energy and the velocity of the daughter nuclei.

Homework Equations


Alpha decay
##Q = (m_X-m_X'-m_\alpha)c^2##
Kinetic energy
##T_\alpha = \frac{Q}{1+\frac{m_\alpha}{m_{X'}}}##
##m(^{230}Th) = 230.033128u##
##m(^{226}Ra) = 226.025303u##
##m(^4He) = 4.002603##
##m(\alpha) = 4.00150618u##
##1 u = 931.502 MeV/c^2##

The Attempt at a Solution


I'm not sure if (or why?) I should use the mass of the ##\alpha## particle or the mass of ##^4 He## in the formula. The answer seem to use ##^4 He## but the difference between them is quite large. Is using the ##\alpha## value more "correct"?
Using the mass of the ##\alpha## particle as the mass of ##^4 He## the ##Q## value is
##Q_{^4He} \approx (230.033128-226.025303-4.002603)\cdot 931.502 MeV \approx 4.77 MeV##. (This is the same as the answer to the question)
Using the mass of the ##\alpha## particle
##Q_\alpha \approx (230.033128-226.025303-4.00150618)\cdot 931.502 MeV \approx 5.79 MeV##

The kinetic energy is
##T_{X'} = Q_{^4He}(1-\frac{1}{1+\frac{m(^4He)}{m_{X'}}}) \approx 0.083 MeV##
Assuming the velocity is ##<<c## the velocity is then
##v = \sqrt{\frac{T_X'}{m_{X'}}} \approx c\sqrt{\frac{0.083MeV}{226.025303\cdot 931.502 MeV}} \approx 1.88\cdot 10^5 m/s##.
However the answer key says ##v= 2.66\cdot 10^5 m/s##. Is the difference from relativistic effects? The speed doesn't seem large enough for this to matter.

I also tried to do it using relativistic kinetic energy
##T = (\gamma-1)m_{X'}c^2 \Longrightarrow \gamma = 1+ \frac{T}{m_{X'}c^2} \approx 1+ \frac{0.083}{226.025403\cdot 931.502}\approx 1.00000039##
So the velocity is
##v = c\sqrt{1-1/\gamma} \approx 1.88 \cdot 10^5m/s##
so the difference doesn't seem to be from this.
 
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  • #2
Incand said:
Assuming the velocity is <<c<<cv=T′XmX′≈c√0.083MeV226.025303⋅931.502MeV≈1.88⋅105m/sv=TX′mX′≈c0.083MeV226.025303⋅931.502MeV≈1.88⋅105m/sv = \frac{T_X'}{m_{X'}} \approx c\sqrt{\frac{0.083MeV}{226.025303\cdot 931.502 MeV}} \approx 1.88\cdot 10^5 m/s.

check the expression for v;
you are writing
v= K.E./mass but v= sqrt (2.K.E./mass) but in numbers you put in a sqrt also but factor of 2 is missing.
 
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  • #3
Thanks! That fixes the velocity! Do you(or anyone else?) happen to have any idea about the usage of the ##^4He## mass compared to ##\alpha##?
 
  • #4
Incand said:
I'm not sure if (or why?) I should use the mass of the αα\alpha particle or the mass of 4He4He^4 He in the formula. The answer seem to use 4He4He^4 He but the difference between them is quite large. Is using the αα\alpha value more "correct"?
Using the mass of the αα\alpha particle as the mass of 4He4He^4 He the QQQ value is
Q4He≈(230.033128−226.025303−4.002603)⋅931.502MeV≈4.77MeVQ4He≈(230.033128−226.025303−4.002603)⋅931.502MeV≈4.77MeVQ_{^4He} \approx (230.033128-226.025303-4.002603)\cdot 931.502 MeV \approx 4.77 MeV. (This is the same as the answer to the question)
regarding this you must see the details of Q value calculation;

The Q value of a nuclear reaction is the difference between the sum of the masses of the initial reactants and the sum of the masses of the final products.
This is also the corresponding difference of the binding energies of the nuclei (not per nucleon), since nucleon number is conserved in a reaction. The masses may be provided in a table of mass excesses (∆M(A, Z)), which is the value of M(A, Z) − Amu (usually in MeV), but relative to the corresponding number for the isotope 12C.
see
http://www.astro.princeton.edu/~burrows/classes/403/nucl.masses.fusion.pdf
 
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  • #5
I think I finally understand now. Since the electron masses cancel in the equation I can use either atomic or nuclear masses but I have to be consistent and use the same for all the mass values.
 
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  • #6
I think I finally understand now. Since the electron masses cancel in the equation I can use either atomic or nuclear masses but I have to be consistent and use the same for all the mass values.
@Incand

No you have not understood correctly- in alpha decay the electron mass has to be accounted for.
see below;

Take care to determine which mass you use from tables in the literature.
Generally, the binding energy of the electrons in the atom (∼eVs to keVs) are too small to compete
with the nuclear terms (∼MeVs) and you needn’t worry,
but for nuclei with the largest atomic numbers the atomic correction can be ∼50-100 keV.
In a strong or electromagnetic nuclear reaction, since total nuclear charge is conserved, the Zme drops out when determining Q values. For a weak interaction, since the nuclear charge changes, the extra me must be accounted for explicitly.
 
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  • #7
drvrm said:
In a strong or electromagnetic nuclear reaction, since total nuclear charge is conserved, the Zme drops out when determining Q values. For a weak interaction, since the nuclear charge changes, the extra me must be accounted for explicitly.
Thanks! You just helped me understand a big difference between ##\alpha##- and ##\beta##-decay and why there's extra electrons to account for in ##\beta^+##-decay.
 
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What is an alpha decay?

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, which is composed of two protons and two neutrons. This process reduces the atomic number of the nucleus by two and the mass number by four.

What is the Q-value of an alpha decay?

The Q-value of an alpha decay is the difference in energy between the initial and final states of the decay. It represents the energy released during the decay and is calculated by subtracting the mass of the parent nucleus from the combined mass of the daughter nucleus and the alpha particle.

What is recoil energy in alpha decay?

Recoil energy is the kinetic energy that the daughter nucleus acquires during an alpha decay. This energy is a result of the conservation of momentum, as the alpha particle carries away some of the momentum from the parent nucleus during the decay process.

How is the momentum of an alpha particle related to recoil energy?

The momentum of an alpha particle is directly related to the recoil energy, as the greater the momentum of the alpha particle, the greater the recoil energy of the daughter nucleus. This relationship is described by the equation: momentum = mass x velocity.

How are the Q-value and recoil energy of an alpha decay related?

The Q-value and recoil energy of an alpha decay are both related to the energy released during the decay process. The Q-value represents the total energy released, while the recoil energy represents the energy that goes into the motion of the daughter nucleus. Therefore, the Q-value is equal to the sum of the recoil energy and the energy carried away by the alpha particle.

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