Question

The radio 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

Answer 1

Answer: (B)

According to Planck's quantum theory, a source of radiation emits energy in the form of photons, which travel in straight line.
The energy 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
$ \bigg [ \because v = \frac{ c }{ \lambda } = \frac{ 3 \times 10^8 }{ 1500} = 2 \times 10^5 \, Hz \bigg ] $
$ \therefore E = 1.32 \times 10^{ - 28 } $ J