Question

A watt is a unit of energy per unit time, and one watt (W) is equal to one joule per second ( J⋅s−1) . An 80.0 W incandescent lightbulb produces about 8.00% of its energy as visible light. Assuming that the light has an average wavelength of 510.0 nm, calculate how many such photons are emitted per second by an 80.0 W incandescent lightbulb.

Answer 1

Answer:

1.64 x 10⁻¹⁹ per sec

Explanation:

P = Power of the bulb = 80 W

t = time

λ = wavelength of the light = 510 nm = 510 x 10⁻⁹ m

c = speed of light = 3 x 10⁸ m/s

n = number of photons

E = energy produced by the bulb as light

Energy produced by the bulb as light is given as

E = (0.08) Pt

$\frac{nhc}{\lambda } = (0.08) Pt$

$\left ( \frac{n}{t} \right ) \frac{hc}{\lambda } = (0.08) P$

$\left ( \frac{n}{t} \right ) \frac{(6.63\times 10^{-34})(3\times 10^{8})}{510\times 10^{-9} } = (0.08) (80)$

$\frac{n}{t}$ = 1.64 x 10⁻¹⁹ per sec

Answer 2

Answer:

there are 1.64×10¹⁹ photons being emitted every second.

Explanation:

The 80W light bulb gives off 80 Joules a second. 8% of 80 J = 6.4 Joules.

So every second 6.4 joules of light energy is given off.

h = 6.626068*10^-34

.....................(planck's constant )

c = 2.99792458*10^8....................(the speed of light)

L=510nm= 5.1*10^-7 metres.

............(wavelength)

Now calculate the energy of a single photon with wave length 5.1*10^-7 metres.

E = hc / L

E = (6.626068*10^-34)*(2.99792458*10^8) / (5.1*10^-7)

E = 3.8949906*10^-19 J

So each photon has 3.8949906*10^-19 J of energy

.

As  total energy is 6.4 J.

So the number of photons is =6.4 / (3.8949906*10^-19) = 1.64×10¹⁹

So there are 1.64×10¹⁹ photons being emitted every second.