Answer
See the detailed answer below.
Work Step by Step
a)
In the first reaction, we got 4.03 MeV.
Thus,
$$\dfrac{Q}{m}=\rm \left(\dfrac{4.03\;MeV}{2[2.014102]\;\color{red}{\bf\not}u}\right)\left(\dfrac{1\;\color{red}{\bf\not}u}{1.67\times 10^{-27}\;\color{red}{\bf\not}kg}\right)\left(\dfrac{1\;\color{red}{\bf\not}kg}{1000\;g}\right)$$
$$\dfrac{Q}{m}=\color{red}{\bf 6.0\times10^{23} }\;\rm MeV/g$$
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In the second reaction, we got 3.27 MeV.
Thus,
$$\dfrac{Q}{m}=\rm \left(\dfrac{3.27\;MeV}{2[2.014102]\;\color{red}{\bf\not}u}\right)\left(\dfrac{1\;\color{red}{\bf\not}u}{1.67\times 10^{-27}\;\color{red}{\bf\not}kg}\right)\left(\dfrac{1\;\color{red}{\bf\not}kg}{1000\;g}\right)$$
$$\dfrac{Q}{m}=\color{red}{\bf 4.86\times10^{23} }\;\rm MeV/g$$
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In the third reaction, we got 17.59 MeV.
Thus,
$$\dfrac{Q}{m}=\rm \left(\dfrac{17.59 \;MeV}{[2.014102+3.016049]\;\color{red}{\bf\not}u}\right)\left(\dfrac{1\;\color{red}{\bf\not}u}{1.67\times 10^{-27}\;\color{red}{\bf\not}kg}\right)\left(\dfrac{1\;\color{red}{\bf\not}kg}{1000\;g}\right)$$
$$\dfrac{Q}{m}=\color{red}{\bf 2.1\times10^{24} }\;\rm MeV/g$$
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b) We know that uranium fission gives 200 MeV;
Thus,
$$\dfrac{Q}{m}=\rm \left(\dfrac{200\;MeV}{235\;\color{red}{\bf\not}u}\right)\left(\dfrac{1\;\color{red}{\bf\not}u}{1.67\times 10^{-27}\;\color{red}{\bf\not}kg}\right)\left(\dfrac{1\;\color{red}{\bf\not}kg}{1000\;g}\right)$$
$$\dfrac{Q}{m}=\color{red}{\bf 5.1\times10^{23} }\;\rm MeV/g$$
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The ratio for the three cases is given by:
For the first reaction:
$$\dfrac{5.1\times10^{23}}{6.0\times10^{23} }=\color{red}{\bf0.85}$$
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For the second reaction:
$$\dfrac{5.1\times10^{23}}{4.86\times10^{23} }=\color{red}{\bf 1.049}$$
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For the second reaction:
$$\dfrac{5.1\times10^{23}}{2.1\times10^{24} }=\color{red}{\bf 0.243}$$
