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Standard XII

Chemistry

Parameters of Nucleus

Assuming the ...

Question

Assuming the radius of nucleus to be equal to $R = 1.3 A^{1 / 3} \times 10^{- 15} m$ , where $A$ is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon $= 1.67 \times 10^{- 27} k g$ .

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Solution

For a nucleus having mass number $A$ that is total nucleons $A$
the radius is $R = 1.3 A^{1 / 3} \times 10^{- 15} m$
So volume will be $V = \frac{4 π}{3} (1.3)^{3} A \times 10^{- 45} m^{3} = 9.2 A \times 10^{- 45} m^{3}$
Density $= m a s s / v o l u m e = \frac{A \times 1.67 \times 10^{- 27} k g}{9.2 A \times 10^{- 45}} = 1.8 \times 10^{27} k g / m^{3}$
Number of nucleons per unit volume is $A / V = \frac{A}{9.2 A \times 10^{- 45}} = 1.09 \times 10^{44} / m^{3}$

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Assuming the radius of nucleus to be equal to R=1.3A1/3×10−15m, where A is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon=1.67×10−27kg .

For a nucleus having mass number A that is total nucleons Athe radius is R=1.3A1/3×10−15mSo volume will be V=4π3(1.3)3A×10−45m3=9.2A×10−45m3Density=mass/volume=A×1.67×10−27kg9.2A×10−45=1.8×1027kg/m3Number of nucleons per unit volume is A/V=A9.2A×10−45=1.09×1044/m3

Assuming the radius of nucleus to be equal to $R = 1.3 A^{1 / 3} \times 10^{- 15} m$ , where $A$ is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon $= 1.67 \times 10^{- 27} k g$ .