JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 14)
The radius R of a nucleus of mass number A can be estimated by the formula
R = (1.3 $$ \times $$ 10–15)A1/3 m.
It follows that the mass density of a nucleus is of the order of :
(Mprot. $$ \cong $$ Mneut $$ \simeq $$ 1.67 $$ \times $$ 10–27 kg)
R = (1.3 $$ \times $$ 10–15)A1/3 m.
It follows that the mass density of a nucleus is of the order of :
(Mprot. $$ \cong $$ Mneut $$ \simeq $$ 1.67 $$ \times $$ 10–27 kg)
1024 kg m–3
1010 kg m–3
1017 kg m–3
103 kg m–3
Explanation
$$R = (1.3 \times {10^{ - 15}}){A^{{1 \over 3}}}$$
We know, $$m = pV$$
$$ \Rightarrow $$ $$p = {m \over V}$$
$$ \Rightarrow $$ $$p = {{{m_p}A} \over {{4 \over 3}\pi {R^3}}}$$
$$p = {{{m_p}A} \over {{4 \over 3}\pi \times {{(1.3 \times {{10}^{ - 15}})}^3}A}}$$
$$p \approx {10^{17}}kg/m$$
We know, $$m = pV$$
$$ \Rightarrow $$ $$p = {m \over V}$$
$$ \Rightarrow $$ $$p = {{{m_p}A} \over {{4 \over 3}\pi {R^3}}}$$
$$p = {{{m_p}A} \over {{4 \over 3}\pi \times {{(1.3 \times {{10}^{ - 15}})}^3}A}}$$
$$p \approx {10^{17}}kg/m$$
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