Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 2462

A radioactive sample has \(\mathrm{n}_{0}\) active atom at \(\mathrm{t}=\mathrm{o}\), at the rate of disintegration at any time is \(\mathrm{R}\) and the number of atom is \(\mathrm{N}\), then ratio. $(\mathrm{R} / \mathrm{N})\( varies with time \)(\mathrm{t})$ as.

Short Answer

Expert verified
The ratio \(\frac{R}{N}\) is equal to the decay constant λ and remains constant with time. Therefore, the ratio does not vary with time.

Step by step solution

01

Radioactive Decay Law

The radioactive decay law states that the rate of decay of a radioactive substance is proportional to the number of active atoms present at any given time. Mathematically, this can be represented as: \(R = \lambda N\) where R is the rate of decay at time t, N is the number of active atoms at time t, and λ is the decay constant.
02

Exponential Decay Equation

We can also write an expression for N, the number of active atoms at any time t, in terms of the initial number of active atoms n₀: \(N = n_{0} e^{-\lambda t}\) where N is the number of active atoms at time t, n₀ is the initial number of active atoms, λ is the decay constant, and t is time.
03

Express R in terms of N

We can write the rate of decay R in terms of the number of active atoms N and the initial number of active atoms n₀ using the radioactive decay law: \(R = \lambda N\) Substitute the expression for N from Step 2: \(R = \lambda (n_{0} e^{-\lambda t})\)
04

Calculate R/N as a function of t

From the expression for R in Step 3, divide by N to find the ratio R/N as a function of time t: \(\frac{R}{N} = \frac{\lambda (n_{0} e^{-\lambda t})}{(n_{0} e^{-\lambda t})}\) Simplify, by canceling the common terms (n₀ and e⁽⁻ᴸᵗ⁾): \(\frac{R}{N} = \lambda\)
05

Conclusion

The ratio R/N is equal to the decay constant λ and remains constant with time. Therefore, the ratio does not vary with time.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The nucleus at rest disintegrate into two nuclear parts which have their velocities in the ratio \(2: 1\) The ratio of their nuclear sizes will be (A) \(2^{(1 / 3)}: 1\) (B) \(1: 2^{(1 / 3)}\) (C) \(3^{(1 / 2)}: 1\) (D) \(1: 3^{(1 / 2)}\)

The radio of minimum to maximum wave length in Balmer series is (A) \((1 / 4)\) (B) \((5 / 36)\) (C) \((3 / 4)\) (D) \((5 / 9)\)

starting with a sample of Pure \({ }^{66} \mathrm{cu},(7 / 8)\) of it decays into \(\mathrm{Zn}\), 15 minutes. The half life the sample is (A) \(5 \mathrm{~min}\) (B) \(7.5 \mathrm{~min}\) (C) \(10 \mathrm{~min}\) (D) \(15 \mathrm{~min}\)

Read the following question and choose correct Answer form given below. (A) Both assertion and reason are true. Reason is the correct explanation of the Assertion. (B) Both assertion and reason are true. Reason is not correct explanation of the assertion. (C) Assertion is true but reason is false. (D) Assertion is false and Reason are true. (i) Assertion :- In a radio-active disintegration, an electron is emitted by nucleus. Reason :- electron are always Present in-side the nucleus. (ii) Assertion :- An electron and Positron can annihilate each other creating Photon Reason:- Electron and Positron form a Particle and anti Particle pair. (iii) Assertion:- An isolated radioactive atom may not decay at all what ever be its half time Reason:- Radioactive decay is a statistical Phenomena. (iv) Assertion:- Fragment Produced in the fission of \(\mathrm{u}^{235}\) are active Reason:- The fragments have abnormally high Proton to neutron ratio

It the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus is $3.6 \mathrm{fm}\( the radius of \){ }^{125}{ }_{52} \mathrm{Te}$ nucleus is nearly equal to (A) \(8 \mathrm{fm}\) (B) \(6 \mathrm{fm}\) (C) \(4 \mathrm{fm}\) (D) \(5 \mathrm{fm}\)
See all solutions

Recommended explanations on English Textbooks

Prosody

Read Explanation

Semiotics

Read Explanation

Sociolinguistics

Read Explanation

International English

Read Explanation

Essay Prompts

Read Explanation

Free Response Essay

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free