Q64PE. - College Physics (Urone)

Question

The Galileo space probe was launched on its long journey past several planets in 1989, with an ultimate goal of Jupiter. Its power source is\(11.0{\rm{\;}}\,{\rm{kg}}\)of \(^{{\rm{238}}}{\rm{Pu}}\), a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the\({\rm{5}}{\rm{.59}}\,{\rm{MeV\alpha }}\)particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of\(^{{\rm{238}}}{\rm{Pu}}\)is\(87.7\)years.

1. What was the original activity of the\(^{{\rm{238}}}{\rm{Pu}}\)in Becquerel? 
2. What power was emitted in kilowatts? 
3. What power was emitted\(12.0\)y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping \({\rm{\gamma }}\).

Step-by-Step Solution

Answer

1. The original activity of the \(^{{\rm{238}}}{\rm{Pu}}\) in Becquerel is \(R = 6.95 \times {10^{15}}\,{\rm{Bq}}\).                                    
2. The power was emitted in kilowatts is \({P_0} = 6.22\,{\rm{kw}}\).
3. The power was emitted \({\rm{12}}{\rm{.0}}\) y after launch is \(P = 5.65\,{\rm{kW}}\).

Concept Introduction

The following is the relationship between activity, half-life, and the number of atoms:

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

\({\rm{Where }}{{\rm{t}}_{{\rm{1/2}}}}{\rm{ = half life,R = activity and N = number of  atoms}}{\rm{.}}\)

Find the original activity

1. The number of nuclei is calculated using the formula\(N = n{N_A}\),

Where \(n = \frac{m}{M}\) is the number of moles.

Now, in the preceding equation, insert in the values of \(m,M\) , and \({N_A}\), and solve for the value of \({\rm{N}}\).

\(N{\rm{ }} = \frac{{(11\;\,{\rm{kg}})(1000\;\,{\rm{g}})}}{{238\;\,{\rm{g/mol}}}}\left( {6.023 \times {{10}^{23}}\,{\rm{atoms /mol}}} \right)\\ = 2.78 \times {10^{25}}\,{\rm{atoms }}\ end \)

Now, in the previous equation, enter in the values of \({\rm{N}}\) and \({{rm{t}}_{{\rm{1/2}}}}\)and solve for \({\rm{R}}\).

\(R = \frac{{0.693 \times 2.78 \times {{10}^{25}}}}{{(87.7\,{\rm{y}})}}\\ = \frac{{0.693 \times 2.78 \times {{10}^{25}}}}{{(87.7\,{\rm{y}})\left( {3.16 \times {{10}^7}\,{\rm{s}}} \right)}}\\ = 6.95 \times {10^{15}}\,{{\rm{s}}^{{\rm{ - 1}}}}\\ = 6.95 \times {10^{15}}\,{\rm{Bq}}\\R = 6.95 \times {10^{15}}\,{\rm{Bq}}\ end\)

Therefore, the original activity of the \(^{{\rm{238}}}{\rm{Pu}}\) in Becquerel is \(R = 6.95 \times {10^{15}}\,{\rm{Bq}}\).                                           

 

Find power was emitted in kilowatts

1. We know that power is defined as the rate of change of energy and is calculated as follows:

\({P_0} = \frac{{\Delta E}}{{\Delta t}}\)

Where\({\rm{\Delta E = }}\)energy change and\({\rm{\Delta t}}\) equals time is\(\frac{{{t_{\frac{1}{2}}}}}{{0.693}}\).

Fill in the missing value and solve for \({{\rm{P}}_{\rm{0}}}\).

\({P_0} = \frac{{6.95 \times {{10}^{15}} \times (5.59\,{\rm{MeV}})\left( {{{10}^6}\,{\rm{eV}}} \right)\left( {1.6 \times {{10}^{ - 19}}\,\;{\rm{J/eV}}} \right)}}{{1000}}\\ = 6.22\,{\rm{kw}}\\{P_0} = 6.22\,{\rm{kw}}\ end\)

Therefore, power was emitted in kilowatts is \({P_0} = 6.22\,{\rm{kw}}\).

Find the power was emitted

1. The following is the relationship between starting power and power after time t: 

\(P = {P_0}{e^{ - \lambda t}}\)

Where, \({{\rm{P}}_{\rm{0}}}{\rm{ = }}\)the initial power and \({\rm{P = }}\)the ultimate power. \(\lambda  = \frac{{0.693}}{{{t_{1/2}}}}\)and \({\rm{t = time}}\)\({\rm{t = time Now plug in the required value and solve for P}}\).

\(P = \left( {6.22\,{\rm{kW}}} \right){e^{\frac{{ - 0.693(12\,{\rm{y}})}}{{87.7\,{\rm{y}}}}}}\\ = 5.65\,{\rm{kW}}\\P = 5.65\,{\rm{kW}}\ end\)

Therefore, the power was emitted \({\rm{12}}{\rm{.0}}\) y after launch is \(P = 5.65\,{\rm{kW}}\).