# IIT JEE past question Kinetics 9

^{238}_{92}U is radioactive and it emits alpha and beta particles to form ^{206}_{82}Pb.
Calculate the number of alpha and beta particles emitted in this conversio. An ore of ^{238}_{92}U is found to contain ^{206}_{82}Pb and ^{238}_{92}U in the wieght ratio 0.1:1.

The half life period of ^{238}_{92}U is 4.5x109 years.
Find the age of the ore.

1
JOHNCENA IS BACK ·

8 ALPHA
6 BETA

AGE=7.12*10TO THE POWER OF 8

1
JOHNCENA IS BACK ·

IS IT CORRECT OR NOT.PLEASE TELL.

1
sriraghav ·

It is 8 alpha and 6 beta

1
sriraghav ·

oops didnt see cena's post....

1
JOHNCENA IS BACK ·

nishant sir,

1
Pavithra Ramamoorthy ·

age is it 1.5 *1010years(approx)?????

62
Lokesh Verma ·

yes :)

1
JOHNCENA IS BACK ·

sir,
u said yes to wat ramku's ans. or to mine.

62
Lokesh Verma ·

i think ramkumar made some mistake!

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Pavithra Ramamoorthy ·

@john....pls tell me d formula r method u solved.....

1
voldy ·

see N=Noe-Î»t gives the no of remianing U atoms so the no of U atoms converted to Pb is No-N

given ratio No-N / N

then solve.

1
Pavithra Ramamoorthy ·

:-)

1
JOHNCENA IS BACK ·

using formula for t1/2 find rate constant and substitute values as given by srinath bhaiya

1
Optimus Prime ·

U238/Pb206= 3/1

N/N-N0=3/1

N/N0=3/4

but as N=N0 e-\lambdat=4/3

\lambdat= log 3(4/3)

tlog e2/T= log e(4/3) as \lambda=loge2/T

t=1.86x109 years

1
Optimus Prime ·