JEE Advanced 2013 · Paper 1 · Physics Q17

← JEE Advanced 2013 · Paper 1

PhysicsInteger+4 / -1Question 17 of 60

A freshly prepared sample of a radioisotope of half-life 1386 s has activity $10^3$ disintegrations per second. Given that $\ln2=0.693$ , the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is

Your answer

Answer and worked solution

Official answer: 4

Decay constant. $\lambda=\dfrac{0.693}{1386}=5\times10^{-4}$ s⁻¹, so $\lambda t=0.04$ for 80 s.

Fraction decayed. $1-e^{-0.04}\approx0.039$ , i.e. about 4%.

Answer: 4

Topics:Radioactive decay

← Previous question Next question →

JEE Advanced 2013 · Paper 1, Physics question 17