For a radioactive decay reacn. :
X -> Y + q a ( where q is the no. of stoichiometric coefficient )
Following graph was observed :
http://img849.imageshack.us/i/62431305.png/

If the ratio of the volume of the emitted helium gas at different time t1 and t2 be 1/2 . Find the value of q.

1
kunl ·

is Y =Helium?

6
AKHIL ·

:)

actually woh a ki jagah alpha hai

sorry for that...
:P

1
kunl ·

LOL!....this time i must say for first time in life i was NOT DUMB...i really felt the questuion needed more to decide :P[3]
now let me try my hands on it!

1
kunl ·

wow this turned out to be pretty interesting

but too long see i will give u the final equations try to sole them and hopefully that would yield required answer

equation-(i)k*(t2-t1)=ln[(1-y)/(1-x)]
eqution-(ii)y/x=2
equation-(iii)k*a*q-k*V=dV/dt.......limits of integration are for V from qax to qay and for time from t1 to t2!

this would yield the answer as far as i think!

6
AKHIL ·

wud u like to xplain a little plzz??
thoda samjha to de yaar!!

1
kunl ·

natural method use kiya maine when nothing works we start from first principles
take initial concentration a and do all the lengthy mole concept wala jhamela at t1 and t2 then write it in terms of the equations u will get similar equations (if i did not do any mistake)! or may be even if u get different equations u might have assumed something different...but assumption is assumption and it is bound to get cancelled [3]so no worries!

6
AKHIL ·

arey got it finally!!

:)

1
kunl ·

explain me too [1]

6
AKHIL ·

:D

ok i did it this way

X -> Y + q \alpha

at t1 a-x x qx
at t2 a-y y qy

at t1 , a-x = qx

x = (a+1)/2

similarly at t2 y = a/2

vol. of gas is directly proportional to no. of moles of gas

so i found qx / qy

so x/y = 2/q+1

and hence q = 3

hope i m correct
:)

1
kunl ·

at t1 , a-x = qx how?

1
kunl ·

@akhil