The number of half lives elapsed before 93.75 % of a radioactive sample has decayed is

Question

The number of half lives elapsed before 93.75 % of a radioactive sample has decayed is (A) 6 (B) 4 (C) 2 (D) 8

Tardigrade Admin · Accepted Answer

Number of half lives be obtained by using N=N0((1/2))n Here, N=(100-9375) × (N0/100) =(6.25/100) × N0 ∴ ((1/2))n=(625/10000) =((5/10))4=((1/2))4 or n=4

Q. The number of half lives elapsed before $93.75%$ of a radioactive sample has decayed is

6

4

2

8

Number of half lives be obtained by using
$N = N_{0} (\frac{1}{2})^{n}$
Here, $N = (100 - 9375) \times \frac{N _{0}}{100}$
$= \frac{6.25}{100} \times N_{0}$
$∴ (\frac{1}{2})^{n} = \frac{625}{10000}$
$= (\frac{5}{10})^{4} = (\frac{1}{2})^{4}$
or $n = 4$

Q. The number of half lives elapsed before 93.75% of a radioactive sample has decayed is

6

4

2

8

Number of half lives be obtained by using N=N0​(21​)n Here, N=(100−9375)×100N0​​ =1006.25​×N0​ ∴(21​)n=10000625​ =(105​)4=(21​)4 or n=4

Q. The number of half lives elapsed before $93.75%$ of a radioactive sample has decayed is

Number of half lives be obtained by using
$N = N_{0} (\frac{1}{2})^{n}$
Here, $N = (100 - 9375) \times \frac{N _{0}}{100}$
$= \frac{6.25}{100} \times N_{0}$
$∴ (\frac{1}{2})^{n} = \frac{625}{10000}$
$= (\frac{5}{10})^{4} = (\frac{1}{2})^{4}$
or $n = 4$