The counting rate observed from a radioactive source at t = 0 second was N 0 counts per second and at 4 t second 16 N 0 counts per second. The counting rate observed, as counts per second, at 5 t second will be

Δ t = 4 t 16 N 0 = d t d N = N 0 e − λ × 4 t 16 1 = e − 4 λ t 4 ln 2 = 4 λ t λ = t ℓ n 2 [RT-6_12th_27-10-13_P-1_Main]

Start Free Mock LoginAlready registered? Login here

Physics - Modern Physics Question with Solution | TestHub

PhysicsModern PhysicsRadioactivityMedium2 minQB

PhysicsMediumsingle choice

The counting rate observed from a radioactive source at $t = 0$ second was $N_{0}$ counts per second and at 4 t second $\frac{N _{0}}{16}$ counts per second. The counting rate observed, as counts per second, at 5 t second will be

Options:

Answer:

Solution:

$Δ t = 4 t \frac{N _{0}}{16} = \frac{d N}{d t} = N_{0} e^{- λ \times 4 t} \frac{1}{16} = e^{- 4 λ t} 4 ln 2 = 4 λ t λ = \frac{ℓ n 2}{t}$

[RT-6_12th_27-10-13_P-1_Main]

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...