Two radioactive elements X and Y have half-lives of 100 and 50 years respectively. Samples of X and Y initially contain equal number of atoms. What is the ratio of the number of the remaining atoms of X to that of Y after 200 years?

Erudite_Scholar

5 years ago

in 200years
X ----- X/2 ----- X/4
Y ----- Y/2 ----- Y/4 ----- Y/8 ----- Y/16

so ratio f X to Y is

X/4 : Y/16

X/Y = 4/1

same as

4:1

Tofunmi3

6 years ago

It's A
N/No=(1/2)^n
Where
N is the atoms that remains
No is the initial atom
n=t/t½
from the que t=200
for atom X
n=200/100=2
Nx/No=(½)^2=1/4

for atom Y
n=200/50=4
Ny/No=(½)^4
Ny/No=1/16

Nx=No/4
Ny=No/16

Nx:Ny
=No/4:No/16
crossmultiply you'll get 4:1

Also
the longer the half life,the slower the decay

so,Atom X with half life 100 will decay slower leaving a larger no than atom Y
X:Y=
4:1

justgerrard

2 years ago

answer is A

Stock

10 years ago

Pls explain.

Anthony_Akaniyene

3 years ago

Correct option is (D)
As we know,
N=N
0
​
/2
n
no. of particles after n half lives
here,
200 years means 4 half life of X nad 2 half life of Y.
AmountleftofX=
2
4

N
0
​

​
∵n=4 (half-lives)
AmountleftofY=
2
2

N
0
​

​
∵n=2
∴RatioofX:Yleft=
2
4

2
2

​
=
4
1
​

6Tam

1 year ago

As the comments say, the answer is A