A helium-neon laser emits light of wavelength 632.8nm. Calculate the energy of each photon in electron volt.

Options

A. 2.5
B. 1.96
C. 0.53
D. 3.3

Answer

B

Explanation

The energy of each photon is
`E=hv`
Since the wavelength and frequency are related to the speed of light by
`c=vlamda`
`E=(hc)/(lamda)=((6.626xx10^(-34))(3.00xx10^(8)))/(632.8xx10^(-9))`
`=3.14xx10^(-19)J`
`1eV=1.602xx10^(-19)J`
`E=(3.14xx10^(-19))/(1.602xx10^(-19))=1.96 eV`
The number of photons emitted per second is equal to the energy emitted by the laser each second divided by the energy of one photon .
`N=(1.00xx10^(-3))/(3.14xx10^(-19))=3.18xx10^(15)` photon `s^(-1)`
`hc=(6.626xx10^(-34))(3xx10^(8))`
`=1.99xx10^(-25)J m=1.24xx10^(3)eVnm`
If a light of wavelength `lamda` nm is incident , energy of photon, in eV, is `E=(1.24xx10^(3))/(lamda)`