# Light bulb operating at voltage U = 220V and consuming current I = 0.27A. emitted per second n = 1.0 * 10 photons.

**Light bulb operating at voltage U = 220V and consuming current I = 0.27A. emitted per second n = 1.0 * 10 photons. The emission of a light bulb consumes n = 7.0% of its power. determine the radiation wavelength corresponding to the average photon energy.**

U = 220 B.

I = 0.27 A.

t = 1 s.

n = 1 ^ 10.

η = 7%.

s = 3 * 10 ^ 8 m / s.

h = 6.6 * 10 ^ -34 J * s.

λ -?

The power consumption of the lamp Npot is determined by the formula: Npot = U * I, where U is the voltage in the lamp, I is the current strength.

According to the condition: only 7% of this power goes to the emission of photons Nfot = η * Npot / 100%.

The power of photons Nfot is determined by the formula: Nfot = n * E1, where n is the number of photons that flew out in 1 s, E1 is the energy of one photon.

The energy of one photon E1 is determined by the formula: E1 = h * c / λ, where h is Planck’s constant, c is the speed of light, λ is the wavelength of the photon.

n * h * s / λ = η * U * I / 100%.

λ = n * h * s * 100% / η * U * I.

λ = 1 ^ 10 * 6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s * 100% / 7% * 220 V * 0.27 A = 4.76 * 10 ^ -26 m …

Answer: the wavelength of the photon is λ = 4.76 * 10 ^ -26 m.