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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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