   Chapter 31, Problem 83PE

Chapter
Section
Textbook Problem

Construct Your Own ProblemConsider 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.

To determine

A problem with given information and also calculate the values.

Explanation

Given info:

Consider the decay of radioactive substance in the Earths interior. The energy emitted is converted to thermal energy that reaches the earths surface and is radiated away into cold dark space.

Formula used:

Using the energy emitted by per unit cubic centimeter per unit second of earths rock, E=QVEarth.

Here, Q is the heat radiated per unit time and VEarth is volume of the earth.

Calculation:

The interior of earth is emitting radioactive substance with some energy. The heat energy emitted by per unit time from surface is qMeV/s . Determine the activity of earth rock in cubic meter, how much power generated? And also how much power must cross each square meter of the Earths surface if the rate of dissipation of power is equal to generation rate of power.

From above formula,

E=qMeV43πR3

Here q is the energy emitted by per unit time from surface and R is the radius of earth, its value is 6380 Km.

VEarth=43π(6380)3( ( 10 6 cm ) 31Km)VEarth=1.09×1030cm3

Therefore,

E=q1.09×1030MeV/s.cm3

The activity of Earth`s rock is

A=EEa=qMeV/s .cm 31

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