SATHEE: Atoms And Nuclei Question 329

Atoms And Nuclei Question 329

Question: The nuclear radius of $_{8} O^{16}$ is $3 \times 10^{- 15}$ . If an atomic mass unit is $1.67 \times 10^{- 27} k g$ , then the nuclear density is approximately

Options:

A) $2.35 \times 10^{17} g / c m^{- 3}$

B) $2.35 \times 10^{17} k g / c m^{- 3}$

C) $2.35 \times 10^{17} c m^{- 3}$

D) $2.35 \times 10^{17} k g m m^{- 3}$

Show Answer

Answer:

Correct Answer: B

Solution:

For nucleus of $_{8} O^{16}$ Mass $= (16), (1.67 \times 10^{- 27}) k g$

Volume $= \frac{4}{3} π R^{3}$

$= \frac{4}{3} π {(3 \times 10^{- 15})}^{3} m^{3} = 36 π \times 10^{- 45} m^{3}$

Density $= \frac{m a s s}{v o l u m e} = \frac{16 \times 1.67 \times 10^{- 27} k g}{36 π \times 10^{- 45} m^{3}}$ $= 2.35 \times 10^{17} k g m^{- 3}$

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Atoms And Nuclei Question 329

Question: The nuclear radius of 8O16 is 3×10−15 . If an atomic mass unit is 1.67×10−27kg , then the nuclear density is approximately

Options:

Answer:

Solution:

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Question: The nuclear radius of $_{8} O^{16}$ is $3 \times 10^{- 15}$ . If an atomic mass unit is $1.67 \times 10^{- 27} k g$ , then the nuclear density is approximately