Chapter 30, Problem 88A

To determine

To calculate:

Kinetic energy of the recoiling thorium nucleus.

Answer to Problem 88A

Kinetic energy of the recoiling thorium nucleus is $0.20\text{MeV}$ .

Explanation of Solution

Given:

  $\begin{array}{l}m\left(U\right)=232.0372\text{u}\\ m\left(Th\right)=228.0287\text{u}\\ m\left(\alpha particle\right)=4.0026\text{u}\end{array}$

K.E $=5.3\text{MeV}$

Formula Used:

Formula for Kinetic energy,

  $K.E_1=$ kinetic energy $-K.E_2$

Where,

  $K.E_1:$ Kinetic energy of $\text{α}\text{particle}$

  $K.E_2:$ Kinetic energy of recoiling thorium nucleus

Calculation:

Total energy can be calculated by subtracting all the masses,

  $\begin{array}{l}=232.0372\text{u}-228.0287\text{u}-4.0026\text{u}\\ =0.0059\text{u}\end{array}$

Converting the above value of energy into $\text{MeV}$ ,

  $\begin{array}{l}=\left(0.0059\text{u}\right)\times \left(931.49\text{MeV}\right)\\ =5.495791\text{MeV}\end{array}$

Kinetic energy of the recoiling thorium nucleus can be calculated as,

  $K.E_1=$ Total energy $-K.E_2$

Substituting the values in above formula,

  $\begin{array}{l}K.E_1=5.495791\text{MeV}-5.3\text{MeV}\\ K.E_1=0.20\text{MeV}\end{array}$

Conclusion:

Hence,kinetic energy of the recoiling thorium nucleus is $0.20\text{MeV}$ .