We study the isotope 37 87 Rb , which decays into the stable isotope 38 87 Sr with a half-life of 5 × 1 0 10 years. Assume that the sample did not contain any isotope 38 87 Sr before. The study was made for one year.

37 87 Rb ⟶ 37 87 Rb + − 1 0 e + r N Rb = N 0 e − λ t N Sr = N 0 ( 1 − e − λ t ) N Rb N Sr = e − λ t 1 − e − λ t = e λ t − 1 ≈ 1 + λ t − 1 = λ t ] [RT-6_12th_09-11-14]

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PhysicsModern PhysicsRadioactivityMedium2 minQB

PhysicsMediummultiple choice

We study the isotope $_{37}^{87} Rb$ , which decays into the stable isotope $_{38}^{87} Sr$ with a half-life of $5 \times$ $1 0^{10}$ years. Assume that the sample did not contain any isotope $_{38}^{87} Sr$ before. The study was made for one year.

Options:(select one or more)

Answer:

A, C

Solution:

$_{37}^{87} Rb ⟶_{37}^{87} Rb +_{- 1}^{0} e + \overline{r}$

$N_{Rb} = N_{0} e^{- λ t} N_{Sr} = N_{0} (1 - e^{- λ t}) \frac{N _{Sr}}{N _{Rb}} = \frac{1 - e ^{- λ t}}{e ^{- λ t}} = e^{λ t} - 1 \approx 1 + λ t - 1 = λ t]$

[RT-6_12th_09-11-14]

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: QB

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