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Home » The unit of length convenient on the nuclear scale is a fermi: 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :r = r0 A1/3where r is the radius of the nucleus, A its mass number, and r0 is a constant equal toabout, 1.2 f. Show that the rule implies that nuclear mass density is nearly constantfor different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to
about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant
for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
CBSE | Class 11 | NCERT | Physics | Units and Measurements
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to
about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant
for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer:

We know that the expression for Radius of the nucleus is given by –

r = r0 A1/3

where ro = 1.2 f = 1.2 x 10-15 m

Now, assuming that the nucleus is spherical. Volume of nucleus can be given by the expression –
V = 4/3 πr3 

V = 4/3 π [r0 A1/3]3 

V = 4/3 πr03A

Let the mass of nucleus = mA

here, m is the average mass of the nucleon

A represents the number of nucleons

Expression for nuclear mass density is –

Nuclear mass density = Mass of nucleus/Volume of nucleus
= mA/(4/3πr3)

= 3mA/4πr3 

= 3mA/4πr03A

= 3m/4πr03

Substituting m = 1.66 x 10-27  kg  and ro = 1.2 f = 1.2 x 10-15 m in the above equation, we get –

Nuclear mass density = 3 x 1.66 x 10-27 /4 x 3.14 x ( 1.2 x 10-15)3

= 4.98 x 10-27/21. 703 x 10-45

Nuclear mass density = 2.29 x 1017 kg/m3

As a result, the nuclear mass density for a sodium atom is substantially higher than the atomic mass density we found in 2.27.

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