Question

A discharge lamp of 36 watt emits yellow light of
wavelength 300 nm in visible region. The number of
photons emitted by the lamp per second is
(consider, value of h = 6 x 10-34 J-s)


Help! ​

Answer 1

Answer:

Power = 36 watt = 36 J/s

Wavelength = 300 nm = 300 * 10^(-9) m

h = 6 x 10^(-34) J-s

To find :

The number of  photons emitted by the discharge lamp per second .

Solution :

E = nhc / λ

where E is energy,

n is the number of photons per second,

c is the speed of light , λ is wavelength and h is constant .

now ,

     36 = n * 6 * 10^(-34) * 3 * 10^(8) / 300 * 10^(-9)

=>  600 = n * 10^(-17)

=>   n = 6 * 10^(19) photons per second

The number of  photons emitted by the discharge lamp is 6 * 10^(19) photons per second .

Explanation: