alpha particle emitting source?!?!?

Question

i havent got the slightest idea how to answer this:

A smoke detector operates using alpha particles from a radioactive Am 241 source.
A typical detector contains 0.2 milligrams of Am 241. the detector can only function if the mass of Am 241 exceeds 0.15 milligrams. Assuming that the lifetime of Am 241 is 624 years, calculate the operational time of the detector.

also does this mean its halflife is 624 / 2 = 312 years?
i've never seen any question like this before

thanks all who have answered but i still dont understand the thinking behind your answers, could the first 3 answerers elaborate a bit?

thanks all who have answered but i still dont understand the thinking behind your answers, could the first 3 answerers elaborate a bit?

Answer 1

What you are dealing with is exponential decay. Half life and mean life are all related to exponential decay.

624 years is the 'mean' lifetime of Am 241. This is how long an atom of Am 241 will, on average, be in existence. It is related to, but longer than half-life.

Half life is when one half of Am 241 is gone. But some Am 241 hangs around for many, many half-lives. This skews the mean life upwards. Half life is the ln (2) [fyi, ln 2=0.693] of the mean life. As a hypothetical, let's get the half life from the mean life:

Half life = Mean life * ln (2)
(Note the "2," above, is just the inverse of 1/2 -- which is the amount of Am you have at the half life time)
Half life = 624 * 0.693
= 432 years

For exponential decay, see: http://en.wikipedia.org/wiki/Exponential_decay
For mean life see: http://en.wikipedia.org/wiki/Mean_lifetime
For half-life, see: http://en.wikipedia.org/wiki/Half_life

What you are looking for is not half-life, not mean life, but 3/4s life -- when 3/4s of the Am 241 is left, i.e., 15 mgs of Am 241.

The simple way to get this is to multiply the mean life times ln (4/3)

3/4s life = Mean life * ln (4/3)
= 624* 0.288
= 179 years

I hope that, with the input of the other answers, I have made this clearer.

Answer 2

Am 241 has a half-life of 432 years. I have no idea where they got 624 years for a lifetime.

Answer 3

The question is a little ambiguous. "Lifetime" can mean at least two different things. A half-life is the time is takes for half the stuff to decay. A characteristic lifetime (tau) is the time it takes so that only a fraction 1/e of the stuff remains. My best guess would be that the 624 years is a characteristic lifetime.

So if you need 3/4 (amount needed / starting amount) of the stuff to remain, that means that:

1/e ^ (t / tau) = (amount needed / starting amount)

t = tau ln (starting amount / amount needed)

Just to answer your question, if my assumption is correct, the half-life would be tau over natural log of 2.

Answer 4

rigidity on an alpha particle via way of the magnetic field = Bqv B - Magnetic field potential q - can value of the particle v - velocity of the particle rigidity in process the middle together as shifting in a around course = mv^2/r m - Mass v - velocity r - Radius can value of one electron = 6.102 x 10^-19 C Bqv = mv^2/r B x ( 6.102 x 10^-19 x 2 ) x a million.3 x 10^7 = [ 6.6 x 10^-27 x ( a million.3 x 10^7 )^2 ] / 0.23 B = 0.3056 H

Answer 5

You need 3/4 of the Americium to remain (i.,e. 1/2 of a 1/2 life).....

Therefore-:

624 / 2^0.5 = 441.234 years