Question

When a nucleus of 235U undergoes fission, it breaks into two smaller, more tightly bound fragments. Calculate the binding energy per nucleon for 235U and for the fission product 137Cs .

Answer 1

The binding energy per nucleon for 235U and 137Cs is 7.6 MeV and 8.39 MeV.

The minimal amount of energy needed to take a particle out of a system of particles or to break a system of particles down into its component pieces is known as binding energy (BE).

Given nucleus = 235 U and nucleon = 137 Cs.

a) The expression to find BE of 235 U is written as 235U92.

Atomic number (Z) = 92 = number of protons

Mass number (A) = 235 (number of protons + neutrons)

Number of electrons = 235-92 = 143.

Mass of neutron (mₙ) =  1.00866 u

Mass of proton ($m_H$) = 1.00783 u

The binding energy for the nucleus is calculated using the below formula,

$B = (Zm_H + Nm_n- m_{atom})\times\mathrm{931.49\;MeV/u}$

For 235U,

$\begin{aligned}BE&=[(92\times1.00783 u)+(143\times1.00866 u)-235](\mathrm{931.49\;MeV/u})\\&=\mathrm{1783.8\;MeV}\end{aligned}$

The BE per nucleon is given by,

$\begin{aligned}\text{BE per nucleon}&=\frac{\text{total BE of nucleus}}{\text{number of nucleon}}\\&=\frac{1783.8}{235}\\&=\mathrm{7.6\;MeV}\end{aligned}$

b) The expression to find BE of 137Cs is written as 137Cs55.

Atomic number (Z) = 55 = number of protons

Mass number (A) = 137 (number of protons + neutrons)

Number of electrons = 137-55 = 82.

Mass of neutron (mₙ) =  1.00866 u

Mass of proton ($m_H$) = 1.00783 u

The atomic mass of Cs = 137u

For 137Cs,

$\begin{aligned}BE&=[(55\times1.00783 u)+(82\times1.00866 u)-137](\mathrm{931.49\;MeV/u})\\&=\mathrm{1149.3\;MeV}\end{aligned}$

The BE per nucleon is given by,

$\begin{aligned}\text{BE per nucleon}&=\frac{\text{total BE of nucleus}}{\text{number of nucleon}}\\&=\frac{1149.3}{137}\\&=\mathrm{8.39\;MeV}\end{aligned}$

The answers are 7.6 MeV and 8.39 MeV.

To know more about binding energy:

brainly.com/question/10095561

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