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

## 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.

Last edited:

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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.

Incand
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##?

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

Incand
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.

drvrm
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.

Incand
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.

drvrm