Binding energy gain indicates increased stability after nuclear reactions
Step 1: Calculate initial and final binding energies - Initial binding energy \[ B_{\text{initial}} = 242 \times 7.6 = 1839.2 \, \text{MeV}. \] Final binding energy \[ B_{\text{final}} = 2 \times (121 \times 8.1) = 2 \times 980.1 = 1960.2 \, \text{MeV}. \]
Step 2: Calculate the gain in binding energy - Gain in binding energy is given by: \[ \text{Gain} = B_{\text{final}} - B_{\text{initial}} = 1960.2 - 1839.2 = 121 \, \text{MeV}. \]
Final Answer: The total gain in binding energy is 121 MeV.
From a height of 'h' above the ground, a ball is projected up at an angle \( 30^\circ \) with the horizontal. If the ball strikes the ground with a speed of 1.25 times its initial speed of \( 40 \ ms^{-1} \), the value of 'h' is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: