If the total binding energies of _ 1 ^ 2 H, _ 2 ^ 4 He, _ 26 ^ 56 Fe \ and _ 92 ^ 235 U nuclei are 2.22,28.3,492 and 1786 MeV respectively, identify the most stable nucleus of the following.

Correct answer: A. _ 26 ^ 56 Fe aligned B.E_ H &=(2.22)/(2)=1.11 B.E_ He &=(28.3)/(4)=7.08 B.E_ Fe &=(492)/(56)=8.78= maximum B.E_ U &=(1786)/(235)=7.6 aligned (56)/(26) Fe is most stable as it has maximum binding energy per nucleon.

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Physics
Chapter: Nuclear Physics

BITSAT 2012 · Physics · Nuclear Physics

If the total binding energies of _ 1 ^ 2 H, _ 2 ^ 4 He, _ 26 ^ 56...

If the total binding energies of

{ }_{1}^{2} \mathrm{H},{ }_{2}^{4} \mathrm{He},{ }_{26}^{56} \mathrm{Fe} \ \text{ and } { }_{92}^{235} \mathrm{U}

nuclei are 2.22,28.3,492 and 1786 MeV respectively, identify the most stable nucleus of the following.

Options

A ${ }_{26}^{56} \mathrm{Fe}$
B ${ }_{1}^{2} \mathrm{H}$
C ${ }_{92}^{235} \mathrm{U}$
D ${ }_{2}^{4} \mathrm{He}$

Correct answer

A. ${ }_{26}^{56} \mathrm{Fe}$

Step-by-step solution

\begin{aligned} \mathrm{B.E}_{\mathrm{H}} &=\frac{2.22}{2}=1.11 \\ \mathrm{B.E}_{\mathrm{He}} &=\frac{28.3}{4}=7.08 \\ \mathrm{B.E}_{\mathrm{Fe}} &=\frac{492}{56}=8.78=\text{ maximum } \\ \mathrm{B.E}_{\mathrm{U}} &=\frac{1786}{235}=7.6 \end{aligned}

\frac{56}{26} \mathrm{Fe} is most stable as it has maximum binding energy per nucleon.$

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