The wavelength of light emitted from an LED is related to its energy gap by the formula:
\( \lambda = \frac{1240}{E_g} \),
where:
Substitute \( E_g = 1.42 \, \text{eV} \):
\( \lambda = \frac{1240}{1.42} \).
Perform the calculation: \( \lambda = 875 \, \text{nm} \, \text{(approximately)}. \)
Final Answer: 875 nm.
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: