Answer:
The antenna emits [tex]5.85\times 10^{19}[/tex] photons per second.
Explanation:
Given:
Power of the antenna is, [tex]P=32.5\ kW=32.5\times 10^3\ W[/tex]
Frequency of the antenna is, [tex]f=83.8\ MHz=83.8\times 10^6\ Hz[/tex]
Now, energy possessed by 1 photon is given as:
[tex]E=h\cdot f\\Where,h\to \textrm{Planck's constant}[/tex]
Here, [tex]f=83.8\times 10^6\ Hz,h=6.626\times 10^{-34}\ J.s[/tex]. So,
[tex]E=(6.626\times 10^{-34}\ J.s)(83.8\times 10^6\ Hz)=5.5526\times 10^{-26}\ J[/tex]
Also, total energy of the antenna per second is equal to its power which is equal to [tex]32.5\times 10^3[/tex] J.
Now, number of photons emitted each second is given as:
[tex]n=\frac{\textrm{Total energy of the antenna per second}}{\textrm{Energy of 1 photon}}\\n=\frac{32.5\times 10^3\ J/s}{5.5526\times 10^{-26}\ J}\\n=5.85\times 10^{19}\textrm{ photons per sec}[/tex]
Therefore, the antenna emits [tex]5.85\times 10^{19}[/tex] photons per second.
