Let binding energy per nucleon of nucleus is denoted by E_ bn and radius of nucleus is denoted as r . If mass number of nuclei A, B and 64 and 125 respectively then

Correct answer: D. r_ A E_ b n B r=r_ 0 ( A)^ 1 / 3 r increases with increasing A mass number So, r_ A 56 ^ 56 Fe has highest E_ bn value. So, E_ b n for A=64 is larger as compared to E_ bn for nucleus with A=125E_ bnA > E_ bnB

BITSAT 2014

Exam: BITSAT
Year: 2014
Subject: Physics
Chapter: Nuclear Physics

BITSAT 2014 · Physics · Nuclear Physics

Let binding energy per nucleon of nucleus is denoted by E_ bn and...

Let binding energy per nucleon of nucleus is denoted by

\mathrm{E}_{\mathrm{bn}}

and radius of nucleus is denoted as

r

. If mass number of nuclei

\mathrm{A}, \mathrm{B}

and 64 and 125 respectively then

Options

A $r_{A} < r_{B}, E_{b n A} < E_{b n B}$
B $r_{A} < r_{B}, E_{b n A} < E_{b n A}$
C(r_{A}= $\frac{4}{5}$ r_{B}, E_{b n A} \lt E_{b n B})
D $r_{A} < r_{B}, E_{b n A} > E_{b n B}$

Correct answer

D. $r_{A} < r_{B}, E_{b n A} > E_{b n B}$

Step-by-step solution

\mathrm{r}=\mathrm{r}_{0}(\mathrm{ A})^{1 / 3}

r increases with increasing A mass number So,

r_{A}< r_{B}

as mass number of

A

is smaller

\mathrm{E}_{\mathrm{bn}}

decreases with increasing

\mathrm{A}

for

\mathrm{A} > 56{ }^{56} \mathrm{Fe}

has highest

\mathrm{E}_{\mathrm{bn}}

value. So,

E_{b n}

for

A=64

is larger as compared to

\mathrm{E}_{\mathrm{bn}}

for nucleus with

\mathrm{A}=125\mathrm{E}_{\mathrm{bnA}} > \mathrm{E}_{\mathrm{bnB}}

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