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Problem Solutions

The **percentage** of the **light** power **associated** with **photons** with less **energy** than
1.1 eV is 20.6% . **b**. What would the photodiode efficiency be if **all** the **energy** of.

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I found an answer from phys.libretexts.org

6.4: The Compton **Effect** - **Physics** LibreTexts

Nov 5, 2020 **...** The Compton **effect** is the term used **for** an unusual result observed when ...
Describe how experiments with X-rays confirm the particle **nature** of **radiation** ...
Here the **photon's energy** Ef is the same as that of **a light quantum** of ... The **wave**
**relation** that connects frequency f with wavelength λ and speed c ...

For more information, see 6.4: The Compton **Effect** - **Physics** LibreTexts

Given that

Sodium lamp power P = 100 W

Wavelength of the sodium light \lambda = 589 nm = 589* 10^{9} m

Imaginary figure:

(a) What is the energy per photon associated with the sodium light?

Energy of proton E=h\upsilon\ =\ \frac{hc}{\lambda}

Where, h - Planck's constant (6.626*10^{-34}), speed of light c = 3*10^{8} m/s and \upsilon - frequency

E = \frac{6.626 * 10^{-34} * 3*10^{8}}{589* 10^{9}}

E = 3.37*10^{- 19} Joules

Converting into electric volts

E = \frac{3.37*10^{- 19}}{1.6* 10^{-19}}

E = 2.11 eV

Thus, energy per photon associated with the sodium light E = 2.11 eV

(b) At what rate are the photons delivered to the sphere?

The number of photons delivered to the sphere per second = n

Relation between power and energy of photon

P = nE

Number of photons per second n = \frac{P}{E}

n = \frac{100}{3.37*10^{- 19}}

n = 2.99* 10^{20}

Hence, The number of photons delivered to the sphere per second n = 2.99* 10^{20} photons/ sec