Question

A 9 watt bulb emitted monocromatic light of wavelength 400 mm bulb calculate the no.of proton emitted per second by the bulb.​

Answer 1

Answer:

Number of photons emmitted = 5.521 * 10^ -20

Explanation:

Using the formula,

E= nhc / lambda

where , h = 6.626*10^-34

c =3*10^8

lambda =400*10^-9

And E  = 9 watts

So therefore by substitution,

9= n*6.626*10^-34*3*10^8 / 400*10^-9

n= 6.626*10^-34*3*10^8 / 400*10^-9*9

n=5.521 * 10^ -20

Hope this helps..