Question

Energy carried by a part of short-wave infrared ray at 1000 nm wavelength is __________ eV (rounded off to 2 decimal places). \[ h = 6.626 \times 10^{-34}\ {Js}, \quad 1\ {J} = 6.242 \times 10^{18}\ {eV}, \quad c = 3 \times 10^8\ {ms}^{-1} \]

Hint:

To convert energy from wavelength, use \( E = \frac{hc}{\lambda} \), and multiply by \( 6.242 \times 10^{18} \) to convert from joules to electronvolts.

Answer 1

The energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] Convert wavelength to meters: \[ \lambda = 1000\ {nm} = 1000 \times 10^{-9} = 1 \times 10^{-6}\ {m} \] Now calculate energy in joules: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1 \times 10^{-6}} = 1.9878 \times 10^{-19}\ {J} \] Convert joules to electronvolts: \[ E = 1.9878 \times 10^{-19} \times 6.242 \times 10^{18} \approx 1.24\ {eV} \] \[ \boxed{1.24\ {eV}} \]