/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q31.5-55PE What fraction of the 40 K that w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What fraction of the 40 K that was on Earth when it formed \(4.5 \times {10^9}\,{\rm{year}}\) ago is left today?

Short Answer

Expert verified

\(0.0874\)of the original \(^{40}\;{\rm{K}}\) is remaining today.

Step by step solution

01

 Definition of fraction 

A fraction is a portion of a whole or, more broadly, any number of equal parts. In everyday English, a fraction describes the number of parts of a specific size, such as one-half, eight-fifths, or three-quarters.

02

Calculation

Consider the original figure of\(^{40}\;{\rm{K}}\)the current number is\({N_0}\), and the atom is\(N\).

As a result, we have according to the radioactive decay equation.

\(\begin{array}{c}N = {N_0}{e^{ - \lambda t}}\\\frac{N}{{{N_0}}} = {e^{ - \lambda t}}\end{array}\)

The decay constant,\(\lambda \), is defined as follows:

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)

The half-life of the radioactive element is\({t_{1/2}}\).

We now have\(^{40}\;{\rm{K}}\).

\({t_{1/2}} = 1.28 \times {10^9}\,{\rm{y}}\)

And\(t\)is the time

\(t = 4.5 \times {10^9}\,{\rm{y}}\)

As a result, the fraction is

\(\begin{array}{c}\frac{{{N_0}}}{N} = {e^{ - \frac{{0.693}}{{{L_{1/2}}}}t}}\\ = {e^{ - \frac{{0.693}}{{\left( {1.28 \times {{10}^9}\,y} \right)}}\left( {4.5 \times {{10}^9}\,{\rm{y}}} \right)}}\\ = 0.0874\end{array}\)

Therefore

Hence, just \(0.0874\) of the original \(^{40}\;{\rm{K}}\) is remaining today.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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

A \({}^{{\rm{60}}}{\rm{Co}}\) source is labelled \(4.00\,{\rm{mCi}}\), but its present activity is found to be \(1.85 \times {10^7}\,{\rm{Bq}}\). (a) What is the present activity in \({\rm{mCi}}\)? (b) How long ago did it actually have a \({\rm{4}}{\rm{.00mCi}}\) activity?

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for electron capture given in theequation \(_Z^A{X_N} + {e^ - } \to _{Z - 1}^A{Y_{N + 1}} + {v_e}\). To do this, identify the values of each before and after the capture.

Verify that a\(2.3 \times {10^{17}}\,{\rm{km}}\)mass of water at normal density would make a cube\(60\,{\rm{km}}\)on a side, as claimed in Example\({\rm{31}}{\rm{.1}}\). (This mass at nuclear density would make a cube\(1.0\,{\rm{m}}\)on a side.)

\({{\rm{\beta }}^{\rm{ - }}}\)decay of \(^{40}K\) , a naturally occurring rare isotope of potassium responsible for some of our exposure to background radiation.

Construct Your Own Problem

Consider the decay of radioactive substances in the Earth's interior. The energy emitted is converted to thermal energy that reaches the earth's surface and is radiated away into cold dark space. Construct a problem in which you estimate the activity in a cubic meter of earth rock? And then calculate the power generated. Calculate how much power must cross each square meter of the Earth's surface if the power is dissipated at the same rate as it is generated. Among the things to consider are the activity per cubic meter, the energy per decay, and the size of the Earth.

See all solutions

Recommended explanations on Physics Textbooks

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.