The activity of a nuclide is 15millicurie. If its decay constant is 0.005 sec−1, the number of atoms present in it is
11.1 x 1010
111 x 1010
1 x 109
1.1 x 1010
A
1.1 x 1010
B
11.1 x 1010
C
1 x 109
D
111 x 1010
Open in App
Solution
Verified by Toppr
Decay constant =0.005sec−1=λ Activity of nuclei =15×3.7×107dps[∵1millicurie=3.7×107dps] =5.55×108dps
Was this answer helpful?
0
Similar Questions
Q1
The activity of a nuclide is 15millicurie. If its decay constant is 0.005 sec−1, the number of atoms present in it is
View Solution
Q2
If the decay constant of a radioactive nuclide is 6.93 x 10−3sec−1, its half life in minutes is:
View Solution
Q3
The activity of a radioactive nuclide (X100) is 6.023 curie at a certain time ‘t′. If its disintegration constant is 3.7×104s−1 the mass of X after t sec is:
View Solution
Q4
The activity of a radioactive nuclide (X100) is 6.023 curie. If its disintegration constant is
3.7 x 104sec−1, then the mass of X is
View Solution
Q5
The activity of a radioactive nuclide (X100) is 6.023 curie at a certain time ‘t′. If its disintegration constant is 3.7×104s−1 the mass of X after t sec is: