From the relation R = R 0 A 1 / 3 , where R 0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A ).
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(Radius of the nucleus of mass number A) R= R₀A^(1/3) - (i)
(R₀= 1.2 × 10^-15)
Volume of nucleus is proportional to R³ which is proportional to A
Volume of nucleus = 4/3 π R³
( substitute i )
=> 4/3π(R₀A^1/3)³
=> 4/3R₀³A
Density of nucleus = mass / volume of nucleus
=mA/(4/3πR₀³A)
=3m/4πR₀³
The above derived equation shows that the density of nucleus is constant , independent of A for all Nuclei.
Hope my answer is helpful to u.
(R₀= 1.2 × 10^-15)
Volume of nucleus is proportional to R³ which is proportional to A
Volume of nucleus = 4/3 π R³
( substitute i )
=> 4/3π(R₀A^1/3)³
=> 4/3R₀³A
Density of nucleus = mass / volume of nucleus
=mA/(4/3πR₀³A)
=3m/4πR₀³
The above derived equation shows that the density of nucleus is constant , independent of A for all Nuclei.
Hope my answer is helpful to u.
Answered by
0
Answer:
R=r
0
A
1/3
R
3
=r
0
3
A
Density=
Volume
Mass
=
3
4
πR
3
Am
p
=
3
4
πr
0
3
A
Am
p
=
4πr
0
3
3m
p
Hence mass density does not depend on mass (A) or atomic number (z)
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