Q157 / 160 · TS EAMCET 2022 (18 Jul Shift 2) · Physics

Consider a nucleus ${}_{30}^{60}X$ . It's approximate density is (Take $1 amu = 1.6 \times 10^{- 27} kg, R_{o} = 1.2 \times 10^{- 15} m$ )

A
$1.2 \times 10^{18} kg m^{- 3}$
B
$8.5 \times 10^{19} kg m^{- 3}$
C
$3.3 \times 10^{16} kg m^{- 3}$
Correct answer
D
$2.2 \times 10^{17} kg m^{- 3}$

Solution

$Density= \frac{mass}{volume}$
$= \frac{A \times (1.6 \times 10^{- 27})}{\frac{4 π}{3} \times {(R_{0} A^{\frac{1}{3}})}^{3}} = \frac{3}{4 π} \times \frac{(1.6 \times 10^{- 27})}{{(1.2 \times 10^{- 15})}^{3}} = 2.2 \times 10^{17} kg m^{- 3}$

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

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