JAMB Physics Past Questions & Answers - Page 144

No explanation has been provided for this answer.

No explanation has been provided for this answer.

In general, the energy released during nuclear fission is given by the Einstein's mass-energy equation, given by

$E = \Delta MC^2$

Where $\Delta m$ is the mass defect, and C is the speed of light.
Therefore
$\Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100} \times 1.0g \\ = 1.0 \times 10^{-4} \\ = 1.0 \times 10^{-7}kg\\ \text{Energy Released } = \Delta MC^2 \\ = 1.0 \times 10^{-7} \times (3.0 \times 10^8)^2\\ = 1.0 \times 10^{-2} \times 9.0 \times 10^{16} \\ = 9.0 \times 10^9 J$

From the uncertainty principle the product of the position and momentum of a sub-atomic particle is equal or greater than planck's constant.

i.e $\Delta x. \Delta p \geq h. \\ \text{Thus} 5 \times 10^{-10} \times \Delta p = 6.6 \times 10^{-34} \\ \Delta p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-10}} = 1.32 \times 10^{-24}Ns\\ \text{Therefore Uncertainty in the momentum } = 1.32 \times 10^{-24}Ns$

The speed of sound in a material is given as:

$V = \sqrt{\frac{E}{P}}, Where\\ E = \text{Young's modulus} \\ P = \text{Density }\\ \text{Therefore } V = \sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}} = 5091.75\\ = 5.1 \times 10^3MS^{-1}$