Use the given masses to calculate the amount of energy released by the following reaction:

98

252

Cf→

42

102

Mo+

56

147

Ba+3(

0

1

n) \begin{tabular}{|l|l|} \hline Californium 252 \\ 4.185815×10

−25

kg & Molybdenum 102 1.692220×10

−25

kg \\ \hline \end{tabular} \begin{tabular}{|l|} \hline Barium 147 \\ 2.439856×10

−25

kg \end{tabular} \begin{tabular}{l} Neutron \\ 1.67490×10

−27

kg \\ \hline \end{tabular} [3] 7. An antiproton (p

−

)slows down as it passes through a magnetic field as shown. Draw the track of the antiproton.

To compute the energy released by the above reaction, we must first estimate the mass change and then use Einstein's mass-energy equivalency formula, E = mc².

Here, it is given that:

Mass of Cf-252 = 4.185815×10⁻²⁵ kg

Mass of Mo-102 = 1.692220×10⁻²⁵ kg

Mass of Ba-147 = 2.439856×10⁻²⁵ kg

Mass of neutron = 1.67490×10⁻²⁷ kg

Δm = (Mass of Cf-252) - (Mass of Mo-102 + Mass of Ba-147 + 3 * Mass of neutron)

Δm = (4.185815×10⁻²⁵ kg) - (1.692220×10⁻²⁵ kg + 2.439856×10⁻²⁵ kg + 3 * 1.67490×10⁻²⁷ kg)

Δm = 2.439819×10⁻²⁵ kg

E = (2.439819×10⁻²⁵ kg) * (3.00×10⁸ m/s)²

Calculating this:

E ≈ 6.95×10¹⁴ joules

Thus, the energy released by the given reaction is approximately 6.95×10¹⁴ joules.

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