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Atoms and Nuclei question

2020 · 4 Sep · Shift 2 · Q61

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Atoms and Nuclei
2020 · 4 Sep · Shift 2 · Q61

Atoms and Nuclei question

2020 · 4 Sep · Shift 2 · Q61

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1

Find the Binding energy per neucleon for

{}_{50}^{120}Sn

. Mass of proton mp = 1.00783 U, mass of neutron mn = 1.00867 U and mass of tin nucleus mSn = 119.902199 U. (take 1U = 931 MeV)

A
9.0 MeV
B
8.5 MeV
C
8.0 MeV
D
7.5 MeV

View written solutionFree

Correct answer: B

Given data

For ${}_{50}^{120}\text{Sn}$ $_{50}^{120} Sn$ :
- Atomic number: $Z = 50$
- Mass number: $A = 120$
- Number of neutrons: $N = A-Z = 120-50 = 70$
Mass of proton: $m_p = 1.00783\,\text{u}$
Mass of neutron: $m_n = 1.00867\,\text{u}$
Mass of tin nucleus: $m_{\text{Sn}} = 119.902199\,\text{u}$
$1\,\text{u} = 931\,\text{MeV}$

Mass defect

Mass defect is

\Delta m = Zm_p + Nm_n - m_{\text{nucleus}}

Substitute values:

\Delta m = 50(1.00783) + 70(1.00867) - 119.902199

Now calculate:

50(1.00783) = 50.3915

70(1.00867) = 70.6069

So,

\Delta m = 50.3915 + 70.6069 - 119.902199

\Delta m = 120.9984 - 119.902199 = 1.096201\,\text{u}

Total binding energy

B = \Delta m \times 931

B = 1.096201 \times 931 \approx 1020.56\,\text{MeV}

Binding energy per nucleon

\frac{B}{A} = \frac{1020.56}{120}

\frac{B}{A} \approx 8.50\,\text{MeV}

Match with options

The binding energy per nucleon is approximately

8.5\,\text{MeV}

So the correct option is B.

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