Question

How many photons are emitted per second by a yellow lamp of power 10 watt? (Wavelength of yellow light = 600 nm)​

Answer 1

Answer:

P = 10 W(given)

this means E in 1 sec = 10 J

 % used to convert into photon = 90% (given)

∴ Energy used = 9 J

 Energy used to take out 1 photon = hc/λ = 6.63* 10-34 * 3 * 108/590 * 10-9 = 6.633/590 * 10-17

 No. of photons used = 9/6.63 *3/590 * 10-17 

= 9* 590/6.63 * 3*1017 

= 266.9 * 1017 

= 2.67 * 1019

hope it helps you

please mark me as brainlist please

Answer 2

Concept:

We need to apply the equation, E = nhv where E can also be represented as P × t and v can be represented as c/λ

Given:

The wavelength of Yellow light = 600 nm

Power of lamp = 10 Watt

Find:

We need to determine the number of photons emitted per second by a yellow lamp

Solution:

It has been given to us the wavelength of the yellow lamp, λ = 600 nm = 600 × 10⁻⁹ m

We know, that the energy of photons is given by the equation,

E = nhv where n is the number of photons, h is the Planck's constant and v is the frequency

However, E = P × t and v = c/λ

Therefore, the equation for energy becomes-

P × t = nhc/λ where, P is the power, t is the time, λ is the wavelength, n is the number of photons, c is the speed of light and h acts as plank's constant.

We have, P as 10, t as 1, λ as It has been given to us as the wavelength of the yellow lamp, λ = 600 nm = 600 × 10⁻⁹ m, h as 6.6 × 10⁻³⁴ and c as 3 × 10⁸

Therefore,

10 × 1 = n × 6.62 × 10⁻³⁴ × 3 × 10⁸/600 × 10⁻⁹

10 = n × 0.033 × 10⁻¹⁷

n = 10/0.033 × 10⁻¹⁷

n = 303.0 × 10¹⁷ photons

n = 3.03 × 10¹⁹ photons

Thus, the number of photons emitted per second by a yellow lamp is 3.03 × 10¹⁹ photons.

#SPJ6