Question

If the detector is capturing 3.3×108 photons per second at this wavelength, what is the total energy of the photons detected in one hour?

Answer 1

Answer:

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

Explanation:

The energy carried by electromagnetic radiation is displaced by waves. This energy is not continuous, but is transmitted grouped into small "quanta" of energy called photons. The energy (E) carried by electromagnetic radiation can be measured in Joules (J). Frequency (ν or f) is the number of times a wave oscillates in one second and is measured in cycles / second or hertz (Hz). The frequency is directly proportional to the energy carried by a radiation, according to the equation: E = h.f, (where h is the Planck constant = 6.63 · 10⁻³⁴ J / s).

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. it is expressed in units of length (m). In light and other electromagnetic waves that propagate at the speed of light (c), the frequency would be equal to the speed of light (≈ 3 × 10⁸ m / s) between the wavelength :

$f=\frac{speed of light}{wavelength}$

So:

$E=\frac{h*speed of light}{wavelength}$

In this case, the wavelength is 3.35mm=3.35*10⁻³m and the energy per photon is:

$E=\frac{6.63*10^{-34}*3*10^{8}}{3.35*10^{-3} }$

E=5.93*10⁻²³ $\frac{J}{proton}$

The detector is capturing  3.3*10⁸ photons per second. So, in 1 hour:

$E=5.93*10^{-23} \frac{J}{proton} *3.3*10^{8} \frac{proton}{s} *\frac{60}{1} \frac{s}{minute} *\frac{60}{1} \frac{minute}{hr}$

E=7.04*10⁻¹¹ $\frac{J}{hr}$

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J