Question

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A is the mass number of the nucleus. Estimate the den- sity of the nucleus of 127I (which has a nuclear mass of 2.1 × 10222 g) in grams per cubic centimeter. Compare with the density of solid iodine, 4.93 g cm23.

Answer 1

Density of 127 I =
`\rm 1.79* 10^(14)\ g/cm^3.`

Also,
`\rm Density\ of\ ^(127)I=3.63* 10^(13)* Density\ of\ the\ solid\ iodine.`

Step-by-step explanation:

Given, the radius of a nucleus is given as

`\rm r=kA^(1/3)`.

where,

• 
  `\rm k = 1.3* 10^(-13) cm.`

• A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

`\rm \rho = (M)/(V)=(M)/(\frac 43 \pi r^3)=(M)/(\frac 43 \pi (kA^(1/3))^3)=(M)/(\frac 43 \pi k^3A).`

For the nucleus 127 I,

Mass, M =
`\rm 2.1* 10^(-22)\ g.`

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

`\rm \rho = (2.1* 10^(-22)\ g)/(\frac 43 * \pi * (1.3* 10^(-13))^3* 127)=1.79* 10^(14)\ g/cm^3.`

On comparing with the density of the solid iodine,

`\rm (Density\ of\ ^(127)I)/(Density\ of\ the\ solid\ iodine)=(1.79* 10^(14)\ g/cm^3)/(4.93\ g/cm^3)=3.63* 10^(13).\\\\\Rightarrow Density\ of\ ^(127)I=3.63* 10^(13)* Density\ of\ the\ solid\ iodine.`