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.
Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \( R_p = 1 \, \Omega \) as shown in the figure. An external resistance of \( R_e = 2 \, \Omega \) is connected via the sliding contact.
The current \( i \) is :
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