Question

5% energy of a monocromatic light source is converted into light of wavelength 5600 A if the power of the source be 100 w then how many photones of light will be emitted by it per second

Answer 1

Answer:

1.4*10^19

Explanation:

et us consider that the monochromatic light source emits ‘ n ’ photons per second at a frequency of  v hertz.

For one second ,

P=100watt=n∗h∗v.  

where  h=6.634∗10−34  is called the Planck’s constant .

v=100n∗h.  

Now ,

E=hv=h∗100(n∗h)=100n.  

Now , 5 % of the original particles are changed into 5600 angstroms wavelength .

v′=c5600angstrom=3∗108(5.6∗10−7)  

v′=5.357∗1014Hz.  

E′=hv′=h∗5.357∗1014Hz.  

According to the given condition ,

E′=5  

hv′=5100∗100n=5n.  

Solving ,

n=5hv′=5(6.634∗10−34∗5.357∗1014)  

n=5(35.54∗10−20)  

n=1.4∗1019.  

Thus , the source releases  1.4∗1019 photons per second .