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3 years ago

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The nucleus of the polonium isotope 214 Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleu

s with mass 4 u). Laboratory experiments measure the speed of the alpha particle to be 1.97×107 m/s.
Assuming the polonium nucleus was initially at rest, what is the recoil speed of the nucleus that remains after the decay?

2 answers:

amid [387]3 years ago

6 0

Answer:

368224.29906 m/s

Explanation:

M = Mass of Polonium nucleus = 214 u

V = Velocity of nucleus

m = Mass of Helium nucleus = 4 u

v = Velocity of alpha particle = $1.97\times 10^7\ m/s$

In this system the momentum is conserved

$MV+mv=0\\\Rightarrow MV=-mv\\\Rightarrow 214V=-4\times 1.97\times 10^7\\\Rightarrow V=\dfrac{-4\times 1.97\times 10^7}{214}\\\Rightarrow V=368224.29906\ m/s$

The recoil speed of the nucleus that remains after the decay is 368224.29906 m/s

Maksim231197 [3]3 years ago

4 0

Answer:

368224.29906

Explanation:

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Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

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With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

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