Q123 / 125 · BITSAT 2014 · Physics

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

  • 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
    \(r_{A}=\frac{4}{5} r_{B}, E_{b n A} < E_{b n B}\)
  • Correct answer
    D
    \(r_{A} < r_{B}, E_{b n A} > E_{b n B}\)

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

Hence, the correct answer is option: 'D'.

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