Find the binding energy per nucleon for $^{120}_{50}\text{Sn}$. Mass of proton $m_p = 1.00783\text{ u}$, mass of neutron $m_n = 1.00867\text{ u}$ and mass of tin nucleus $m_{\text{Sn}} = 119.902199\text{ u}$. (Take, $1\text{ u} = 931\text{ MeV}$) Options: (A) 9.0 MeV (B) 7.5 MeV (C) 8.0 MeV (D) 8.5 MeV

### Analyzing the Setup Imagine you are holding a single nucleus of Tin-120. It is a dense, tightly packed sphere containing exactly $50$ protons and $70$ neutrons. If we could magically pull this nucleus apart and weigh each proton and neutron individually on a microscopic scale, the sum of their masses would give us what we call the **theoretical mass**. Let's set up our master equation for this theoretical mass: $$m_{\text{theoretical}} = Zm_p + (A - Z)m_n$$ ### The Master Equation We know the atomic number $Z = 50$ and the mass number $A = 120$. This means we have $50$ protons and $120 - 50 = 70$ neutrons. Substituting the given masses of a single proton and neutron, we get: $$m_{\text{theoretical}} = 50(1.00783\text{ u}) + 70(1.00867\text{ u})$$ Now, we must be incredibly precise. In nuclear physics, rounding off early is a fatal trap! Multiplying and adding these values carefully yields: $$m_{\text{theoretical}} = 50.391500\text{ u} + 70.606900\text{ u} = 120.998400\text{ u}$$ But here is where the universe plays a trick on us. When these $120$ nucleons bind together to form the actual Tin nucleus, the experimental mass is only $119.902199\text{ u}$. A tiny fraction of the mass has simply vanished! This missing mass is known as the **mass defect** ($\Delta m$). Let's calculate exactly how much mass disappeared: $$\Delta m = m_{\text{theoretical}} - m_{\text{experimental}}$$ $$\Delta m = 120.998400\text{ u} - 119.902199\text{ u} = 1.096201\text{ u}$$ ### Final Calculation Where did this missing mass go? Albert Einstein gave us the answer: it was converted entirely into energy to bind the nucleus together! We can find this **Total Binding Energy** by multiplying the mass defect by the conversion factor $931\text{ MeV/u}$. $$\text{BE} = 1.096201\text{ u} \times 931\text{ MeV/u} = 1020.563131\text{ MeV}$$ This is a massive amount of energy, but it represents the energy of the *entire* nucleus. The question asks for the binding energy **per nucleon**, which tells us how tightly each individual particle is held. We find this by dividing the total energy by the mass number $A = 120$. $$\text{BE}_{\text{per nucleon}} = \frac{1020.563131\text{ MeV}}{120} \approx 8.5\text{ MeV}$$ An energy of $8.5\text{ MeV}$ per nucleon is a hallmark of a highly stable, medium-heavy nucleus. The math beautifully aligns with the physical reality, leading us straight to option (d)!

JEE Main 2020

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Animated Solution for Physics - Atoms and Nuclei: Find the binding energy per nucleon for $_{50}^{120} Sn$ . Mass of proton $m_{p} = 1.00783 u$ , mass of neutron $m_{n} = 1.00867 u$ and mass of tin nucleus $m_{Sn} = 119.902199 u$ . (Take, $1 u = 931 MeV$ )

Q: How do I know how many protons and neutrons are in Tin-120?

The symbol $^{120}_{50}\text{Sn}$ gives you all the information. The bottom number (atomic number, $Z$) is 50, which is the number of protons. The top number (mass number, $A$) is 120. The number of neutrons is simply $A - Z = 120 - 50 = 70$.

Q: Why do we multiply by 931 instead of using $E=mc^2$ directly with kilograms and the speed of light?

You absolutely could use $E=mc^2$, but you would have to convert atomic mass units (u) to kilograms, and then the resulting energy from Joules to electron-volts. The factor $1\text{ u} = 931\text{ MeV}$ is a convenient shortcut that does all these conversions for you in one step!

Select Answer:

Visualized Solution

$m_{theoretical} = Z m_{p} + (A - Z) m_{n}$

The theoretical mass of a nucleus is the sum of the masses of its individual protons and neutrons.

$m_{theoretical} = 50 (1.00783) + 70 (1.00867)$

Substitute the given values for $Z = 50$ and $A - Z = 120 - 50 = 70$ .

$m_{theoretical} = 50.391500 + 70.606900 = 120.998400 u$

Calculate the total theoretical mass.

$Δ m = m_{theoretical} - m_{experimental}$

Mass defect is the difference between the theoretical mass and the actual experimental mass of the nucleus.

$Δ m = 120.998400 - 119.902199$

Substitute the calculated theoretical mass and the given experimental mass.

$Δ m = 1.096201 u$

Calculate the mass defect.

$BE = Δ m \times 931 MeV$

Convert the mass defect into Binding Energy using the conversion factor $1 u = 931 MeV$ .

$BE = 1.096201 \times 931 = 1020.563131 MeV$

Calculate the total Binding Energy.

$BE_{per nucleon} = \frac{BE}{A}$

Binding energy per nucleon is the total binding energy divided by the mass number $A$ .

$BE_{per nucleon} = \frac{1020.563131}{120} \approx 8.5 MeV$

Calculate the final value.

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The Sigma Insight: Nucleus and Nuclear Reaction

Solution Diagram

Analyzing the Setup

Imagine you are holding a single nucleus of Tin-120. It is a dense, tightly packed sphere containing exactly

50

protons and

70

neutrons. If we could magically pull this nucleus apart and weigh each proton and neutron individually on a microscopic scale, the sum of their masses would give us what we call the theoretical mass.

Let's set up our master equation for this theoretical mass:

m_{theoretical} = Z m_{p} + (A - Z) m_{n}

The Master Equation

We know the atomic number

Z = 50

and the mass number

A = 120

. This means we have

50

protons and

120 - 50 = 70

neutrons. Substituting the given masses of a single proton and neutron, we get:

m_{theoretical} = 50 (1.00783 u) + 70 (1.00867 u)

Now, we must be incredibly precise. In nuclear physics, rounding off early is a fatal trap! Multiplying and adding these values carefully yields:

m_{theoretical} = 50.391500 u + 70.606900 u = 120.998400 u

But here is where the universe plays a trick on us. When these

120

nucleons bind together to form the actual Tin nucleus, the experimental mass is only

119.902199 u

. A tiny fraction of the mass has simply vanished! This missing mass is known as the mass defect (

Δ m

Let's calculate exactly how much mass disappeared:

Δ m = m_{theoretical} - m_{experimental}

Δ m = 120.998400 u - 119.902199 u = 1.096201 u

Final Calculation

Where did this missing mass go? Albert Einstein gave us the answer: it was converted entirely into energy to bind the nucleus together! We can find this Total Binding Energy by multiplying the mass defect by the conversion factor

931 MeV/u

BE = 1.096201 u \times 931 MeV/u = 1020.563131 MeV

This is a massive amount of energy, but it represents the energy of the entire nucleus. The question asks for the binding energy per nucleon, which tells us how tightly each individual particle is held. We find this by dividing the total energy by the mass number

A = 120

BE_{per nucleon} = \frac{1020.563131 MeV}{120} \approx 8.5 MeV

An energy of

8.5 MeV

per nucleon is a hallmark of a highly stable, medium-heavy nucleus. The math beautifully aligns with the physical reality, leading us straight to option (d)!

Common Traps & FAQs

Why can't we just round off the masses to two decimal places to make the calculation easier?

In nuclear physics, the mass defect is incredibly small, often in the third or fourth decimal place of an atomic mass unit. Rounding off early will completely destroy the accuracy of the mass defect, leading to a drastically wrong binding energy.

How do I know how many protons and neutrons are in Tin-120?

The symbol

_{50}^{120} Sn

gives you all the information. The bottom number (atomic number,

Z

) is 50, which is the number of protons. The top number (mass number,

A

) is 120. The number of neutrons is simply

A - Z = 120 - 50 = 70

Why is the experimental mass always less than the theoretical mass?

When individual nucleons come together to form a nucleus, they release a massive amount of energy to become a stable bound system. According to

E = m c^{2}

, this released energy comes at the cost of mass. Hence, the bound nucleus weighs slightly less than its constituent parts.

Why do we multiply by 931 instead of using

E = m c^{2}

directly with kilograms and the speed of light?

You absolutely could use

E = m c^{2}

, but you would have to convert atomic mass units (u) to kilograms, and then the resulting energy from Joules to electron-volts. The factor

1 u = 931 MeV

is a convenient shortcut that does all these conversions for you in one step!

What does the final value of 8.5 MeV per nucleon physically mean?

The binding energy per nucleon is a measure of nuclear stability. It tells you how much energy, on average, is required to remove a single proton or neutron from the nucleus. A value around 8.5 MeV is typical for medium-heavy nuclei like Tin, indicating it is very tightly bound and stable.