\[ E = \frac{1240}{\lambda (\text{nm})} \, \text{eV} \]
\[ E = \frac{1240}{242} \, \text{eV} \]
\[ E = 5.12 \, \text{eV} \]
\[ E = 5.12 \times 1.6 \times 10^{-19} \, \text{J/atom} \]
\[ E = 8.198 \times 10^{-19} \, \text{J/atom} \]
\[ E = 494 \, \text{kJ/mol} \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32