Answer:
The IR-A is the greatest energy per photon compare to that of IR-B.
Explanation:
Given that,
Wavelength of (IR-B)= 1400 nm
Wavelength of (IR-A)= 700 nm
We need to calculate the energy for IR-B
Using formula of energy
[tex]E=\dfrac{hc}{\lambda}[/tex]
Put the value into the formula
[tex]E=\dfrac{6.62\times10^{-34}\times3\times10^{8}}{1400\times10^{-9}}[/tex]
[tex]E=1.418\times10^{-19}\ J[/tex]
The energy for IR-B is [tex]1.418\times10^{-19}\ J[/tex]
We need to calculate the energy for IR-A
Using formula of energy
[tex]E=\dfrac{hc}{\lambda}[/tex]
Put the value into the formula
[tex]E=\dfrac{6.62\times10^{-34}\times3\times10^{8}}{700\times10^{-9}}[/tex]
[tex]E=2.83\times10^{-19}\ J[/tex]
The energy for IR-B is [tex]2.83\times10^{-19}\ J[/tex]
Hence, The IR-A is the greatest energy per photon compare to that of IR-B.
