Question

A light-emitting diode (LED) connected to a 3.0 Vpower supply emits 440 nm blue light. The current in the LED is 21 mA , and the LED is 41 %efficient at converting electric power input into light power output.

Part A How many photons per second does the LED emit?

Answer 1

Voltage (V) = 3 Volts, Current = 21 mA, Wavelength of blue light (?) = 440 nm

Plank constant (h) = 6.626068 * 10 ^ -34 J-s

c = 3*10^8 m/s

Power of light-emitting diode = Voltage x Current = 3.0 x 21 x 10^-3 = 0.063 W

Efficiency of the LED = 41 %

light power (output) = 0.063 * 0.41 = 0.02583 W

Energy of one photon: Ep = h (c /?) = 6.626068 * 10 ^ -34 * (3*10^8/440*10^-9)

Ep = 0.04518 * 10^-17 J

In 1 sec energy of emmited light is 0.02583 J; or we can say 0.02583 J energy will be emmitted in one sec.

Now,  0.04518 * 10^-17 J energy is emmited by 1 photon;

so, 1 J energy will be emmited by 1/ 0.04518 * 10^-17 photon;

0.02583 J energy will be emmited by (0.02583)*( 1/ 0.04518 * 10^-17) =0.572*10^17.

photons per second emmited = 0.572*10^17.