JEE Main 27-Aug-2021 Shift 1 Solved Paper

Initial number of radioactive nuclei, N0=1010
Half-life, t

1

2

=1‌min=60s
Time, t=30s
We know that, number of nuclei remaining in radioactive decay after time t is given by

N

N0

=(

1

2

)n=(

1

2

)

t

t1∕2

Substituting the values in above relation, we get

N

1010

=(

1

2

)

30

60

=(

1

2

)

1

2

=

1

√2

⇒N=

1

√2

×1010=0.707×1010
⇒N≈7×109
Thus, the radioactive nuclei remaining after 30s is 7×109.