# A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The...

## Question:

A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 21 mA , and the LED is 61 % efficient at converting electric power input into light power output.

How many photons per second does the LED emit?

## Photon Energy

A photon of wavelength {eq}\lambda {/eq} carries an energy of $$E = h \, \frac{c}{\lambda} $$ where {eq}h = 6.626 \times 10^{-34}~\rm J \cdot s {/eq} is the Planck constant and {eq}c {/eq} the speed of light in vacuum.

## Answer and Explanation:

The total power dissipated by the LED is {eq}\rm 3.0~V \cdot 0.021~A = 0.063~W {/eq}. 61% of this power is converted into light, leading to a blue output of {eq}\rm 0.61 \cdot 0.063~W = 0.038~W {/eq}, or 0.038 Joule per second.

The energy of a blue photon at 440 nm is given as $$E = h \, \frac{c}{\lambda} = \rm 6.626 \times 10^{-34}~J \cdot s \cdot \frac{3 \times 10^8~m/s}{440 \times 10^{-9}~m} = 4.52 \times 10^{-19}~J $$ Consequently, the number of photons emitted by the LED per second is $$N = \rm \frac{0.038~J}{4.52 \times 10^{-19}~J} = 8.4 \times 10^{16}~. $$

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from UExcel Physics: Study Guide & Test Prep

Chapter 17 / Lesson 2