Fraction remains after n half lives N0N=(21)n=(21)t/T Given N=eN0⇒eN0N0=(21)5/T or e1=(21)5/T Taking log on both sides, we get log1−loge−1⇒T= T5log21= T5(−log2)=5loge2 Now, let t′ be the time after which activity reduces to half ⇒(21)⇒t′=(21)t′/5loge2=5loge2
The activity of a radioactive sample is measured as N_0 counts per minute at t=0 and N_0 / e counts per minute at t=5 ~min. The time (in minute) at…
NEET UG 2010 — Physics Modern Physics
The activity of a radioactive sample is measured as N0 counts per minute at t=0 and N0/e counts per minute at t=5 min. The time (in minute) at which the activity reduces to half its value is
Held on 30 Apr 2010 · Verified 9 Jul 2026.
Options
- A
loge2/5
- B
loge25
- C
5log102
- D
5loge2
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
NEET UG Modern Physics in other years
More NEET UG Modern Physics questions
A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ ) of their de Broglie wavelengths is: (c is the speed of light)
The output $(\mathrm{Y})$ of the given logic implementation is similar to the output of an/a _____ gate. 
The maximum kinetic energy of photoelectrons from a surface with work function φ = 2 eV when light of λ = 4000 Å falls on it is (hc = 12400 eV·Å):
A full wave rectifier circuit with diodes $\left(D_1\right)$ and $\left(D_2\right)$ is shown in the figure. If input supply voltage $V_{\text {in }}=220 \sin (100 \pi t)$ volt, then at $t=15 \mathrm{msec}$ 
A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius $r$ of the $n^{\text {th }}$ orbit and the particle's speed $v$ in the orbit depend on $n$ as
Other Physics topics for NEET UG
Keep practicing Modern Physics
Work through every NEET UG Modern Physics PYQ, year by year.