Question

A TV center transmits 10 kilowatts of power at 150 MHz. The energy of a photon of the electromagnetic wave is
(A) $6 \times {10^{ - 7}}J$
(B) $6 \times {10^{ - 7}}eV$
(C) $6 \times {10^{ - 17}}J$
(D) $6 \times {10^{ - 17}}eV$

Answer 1

In this question, we need to determine the energy of a photon of the electromagnetic wave such that the television center transmits 10 kilowatts of power at 150 megahertz. In this question frequency of the transmission is given and since we know the value of the Planck's constant so we will find the energy of the transmission, and we know the charge on the photon is equal to $1.6 \times {10^{ - 19}}eV$, so by using this, we will find the energy of a photon.

Complete step by step answer:
Power transmitted by TV center transmits $P = 10kW = 10000W$
Frequency of transmission $v = 150MHz$
We know photon is the smallest particle of an electromagnetic wave where the energy of a photon of the electromagnetic wave is given by the formula
$E = hv - - (i)$
We know that Plank’s constant in an electromagnetic wave is $h = 6.26 \times {10^{ - 34}}{m^2}{kg}{s^{-1}}$
Also, the frequency of transmission is given $v = 150MHz$
Now substitute the values of planck's constant and the frequency in equation (i), we get

$$E = hv \\\ \Rightarrow E= \left( {6.26 \times {{10}^{ - 34}} \times 150 \times {{10}^6}} \right) \\\ \Rightarrow E= 9.94 \times {10^{ - 26}}J - - (ii) \\\$$

Now since we know that
$1J = 1.6 \times {10^{ - 19}}eV$
Hence we can further write equation (ii)

$$E = \dfrac{{9.94 \times {{10}^{ - 26}}}}{{1.6 \times {{10}^{ - 19}}}}eV \\\ \Rightarrow E= 6.21 \times {10^{ - 7}}eV \\\ \therefore\simeq 6 \times {10^{ - 7}}eV \\\$$

Hence the energy of a photon of the electromagnetic wave is $\simeq 6 \times {10^{ - 7}}eV$ and option B is the correct.

Note: Energy of photon of the electromagnetic wave is given by the formula$E = hv$, where $h$is the Plank constant and $v$is the frequency of the electromagnetic wave.