(1): Angle of Minimum Deviation
The refractive index $n$ of the material of the prism is related to the angle of the prism $A$ and the angle of minimum deviation $\delta_m$ by the formula: \[ n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}. \] Substituting the given values: $n = \sqrt{2}$, $A = 60^\circ$. Rearranging the formula: \[ \sin\left(\frac{A + \delta_m}{2}\right) = n \cdot \sin\left(\frac{A}{2}\right). \] Calculating $\sin\left(\frac{A}{2}\right)$: \[ \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ, \quad \sin 30^\circ = \frac{1}{2}. \] Substituting: \[ \sin\left(\frac{A + \delta_m}{2}\right) = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}. \] The angle whose sine is $\frac{\sqrt{2}}{2}$ is $45^\circ$: \[ \frac{A + \delta_m}{2} = 45^\circ. \] Solving for $\delta_m$: \[ A + \delta_m = 90^\circ \implies \delta_m = 90^\circ - 60^\circ = 30^\circ. \] Thus, the angle of minimum deviation is: \[ \boxed{\delta_m = 30^\circ}. \] (2): Angle of Incidence
At the angle of minimum deviation, the angle of incidence $i$ is equal to the angle of emergence. Using the geometry of the prism, the relation between the angle of incidence, the angle of refraction $r$, and the prism angle $A$ is: \[ r = \frac{A}{2}. \] Substituting $A = 60^\circ$: \[ r = \frac{60^\circ}{2} = 30^\circ. \] Using Snell's law at the first face of the prism: \[ n = \frac{\sin i}{\sin r}. \] Rearranging for $i$: \[ \sin i = n \cdot \sin r. \] Substituting: \[ \sin i = \sqrt{2} \cdot \sin 30^\circ = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}. \] The angle whose sine is $\frac{\sqrt{2}}{2}$ is $45^\circ$: \[ i = 45^\circ. \] Thus, the angle of incidence is: \[ \boxed{i = 45^\circ}. \]
A certain reaction is 50 complete in 20 minutes at 300 K and the same reaction is 50 complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. Given: \[ R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}, \quad \log 4 = 0.602 \]