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Chapter 18: Problem 45

The energy released by the formation of a nucleus of \(_{26}^{56} \mathrm{U}\) is \(7.89 \times 10^{-11} \mathrm{J}\) Einstein's equation, \(E=m c^{2},\) to determine how much mass is lost (in kilograms) in this process.

Short Answer

Expert verified
The mass loss in the formation of the given nucleus is \(m = 8.79 \times 10^{-14}\) kg.

Step by step solution

01

Identify the given values

The given values in the problem are the energy \(E = 7.89 \times 10^{-11} \mathrm{J}\) and the speed of light \(c = 3.00 \times 10^{8} \mathrm{m/s}\).
02

Rearrange the equation

Einstein's equation needs to be rearranged to solve for the mass 'm'. Using algebra, the equation becomes \(m = \frac{E}{c^{2}}\).
03

Substitute the values into the equation

By substituting \(E = 7.89 \times 10^{-11} \mathrm{J}\) and \(c = 3.00 \times 10^{8} \mathrm{m/s}\) into the equation, the expression becomes: \(m = \frac{7.89 \times 10^{-11}}{(3.00 \times 10^{8})^{2}}\).
04

Do the calculation

Perform the division to get the answer to the problem. This is the lost mass in kilograms.

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Most popular questions from this chapter

Complete the following nuclear reactions. \begin{equation}a. _{5}^{12} \mathrm{B} \rightarrow_{6}^{12} \mathrm{C}+?\end{equation} \begin{equation}b. _{89}^{225} \mathrm{Ac} \longrightarrow_{87}^{221} \mathrm{Fr}+?\end{equation} \begin{equation}\mathbf{c.}_{28}^{63} \mathrm{Ni} \longrightarrow ?+_{-1}^{0} e\end{equation} \begin{equation}\mathbf{d} \cdot_{83}^{212} \mathrm{Bi} \rightarrow ?+_{2}^{4} \mathrm{He}\end{equation}

The half-life of carbon-14 is 5715 y. How long will it be until only half of the carbon-14 in a sample remains?

Write the nuclear equation for the release of a positron by \(_{54}^{117} \mathrm{Xe} .\)

Explain why charged particles do not pene- trate matter deeply.

What is a nucleon?
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