Chemistry - Radiation

Yuugib

I was doing some previous exam qs on this topic, and i cam across this one that just buffled me

Write and equation for the emission of an alpha particle from americum-241.


Could anybody help me please?)

Cokehead Mother

Well an alpha particle is 4_2. I think americum-241 is 95_241.

You'd have your americum on one side, and your alpha particle on the other side plus something else to balance the equation.

So it's like

241_95 -> 4_2 + 237_93

So you get a protactinium isotope.

To be honest, I don't even do chemistry so take all this with a pinch of salt.

salman85

alpha particle = take 2 from bottom and 4 from top

beta = remove 1 from the bottom and leave top unchanged (i think)

gosh need to revise this for physics

Yuugib

doesnt it supposed to take 4 from the mass? In periodic table americium has a mass of 243..?
i m confused

if it was 241 then it becomes neptunium.. but whats with the 243 in the table? arghh...

kaki

Aye, I find these annoying as (if I remember rightly) they don't give you the atomic number on the exam paper, and the periodic table in the maths tables don't go up to Americium, and you can't predict the resulting particle as it too probably won't be on the table.

I think Americium 241 decays to Neptunium 237, w/ one alpha particle being emitted, though. Could be wrong...

kaki

Americium 243 is the standard form, but Americium 241 is the isotope used in fire alarms?

Yuugib

aw yes yes i see where you comming from
thank you, that sorted me out

ian.f

Don't over complicate is

Americium241_X - 4_2α ---> 237_x-2(Y)

Where x = proton/atomic number and Y is name of undeterminable element

Cokehead Mother

ian.f wrote: »

Don't over complicate is

Americium241_X - 4_2α ---> 237_x-2(Y)

Where x = proton/atomic number and Y is name of undeterminable element

Shammit. Mine was wrong and inconsistent.