Question

How many 860nm photos are contained in a 2.7uJ pulse of monochromatic radiation ?

Answer 1

Answer:
`1.1683* 10^(13)` number of photons.

Step-by-step explanation:

`\text{Energy of photon}=E= (hc)/(\lambda )`

`E= \frac{6.626* 10^(-34)J s* 3* 10^(8)m/s}{860* 10{-9}m}=0.02311* 10^(-17) J`

`\text{Energy of radiation}=2.7\mu J=2.7* 10^(-6)J(1\mu J=10^(-6)J)`

`\text{Energy of radiation}=E* \text{Number of photons}`

`\text{Number of photons}=\frac{\text{Energy of radiation}}{\text{Energy of photon}(E)}=(2.7* 10^(-6)J)/(0.02311* 10^(-17) J)=1.1683* 10^(13)\text{photons}`

`1.1683* 10^(13)` numbers of photons will be contained by
`2.7\mu J` pulse of monochromatic radiation.