The energy gap \( E_g \) of the LED is given as 1.9 eV. The wavelength \( \lambda \) of the emitted light is related to the energy gap \( E_g \) by the equation: \[ E_g = \frac{hc}{\lambda} \] where:
- \( h \) is Planck’s constant (\( 6.626 \times 10^{-34} \, \text{J·s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the emitted light. We also need to convert the energy gap from eV to joules.
Using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), the energy gap in joules is: \[ E_g = 1.9 \, \text{eV} = 1.9 \times 1.602 \times 10^{-19} \, \text{J} = 3.05 \times 10^{-19} \, \text{J} \] Now, using the formula for the wavelength: \[ \lambda = \frac{hc}{E_g} \] Substitute the known values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{3.05 \times 10^{-19}} = 6.54 \times 10^{-7} \, \text{m} = 654 \, \text{nm} \]
Thus, the wavelength of the emitted light is 654 nm. Therefore, the correct answer is: \[ \text{(B) } 654 \, \text{nm} \]
A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?
Two point charges M and N having charges +q and -q respectively are placed at a distance apart. Force acting between them is F. If 30% of charge of N is transferred to M, then the force between the charges becomes:
If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be: