Q. The ratio transmitter operates on a wavelength of 1500 m at a power of 400 kW. The energy of radio photon (in joule) is
A) $1.32 \times 10^{- 24}$ J
B) $1.32 \times 10^{- 28}$ J
C) $1.32 \times 10^{- 26}$ J
D) $1.32 \times 10^{- 32}$ J
Explanation
(b) The energy of photon is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = hv = \frac{hc}{\lambda}$
Hence, energy of radio photon is
$E = 6.6 \times 10^{- 34} \times 2 \times 10^{5}$ J
$\left( \because v = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{1500} = 2 \times 10^{5}\ Hz \right)$
$\therefore\ \ \ \ \ \ \ \ \ E = 1.32 \times 10^{- 28}\ $J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.