Answer:
The energy of 2.00 mol of photons of an infrared radiation 1460 nm is 164.08 kilo Joules.
Explanation:
[tex]E=\frac{h\times c}{\lambda}[/tex]
where,
E = energy of a photon = ?
h = Planck's constant = [tex]6.63\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = [tex]1460 nm^o=1460\times 10^{-9}m[/tex]
Now put all the given values in the above formula, we get the energy of the photons.
[tex]E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{1460\times 10^{-9}m}[/tex]
[tex]E=1.3623\times 10^{-19}J[/tex]
1 mol = [tex]6.022\times 10^{23} [/tex] particles
Then in 2 moles of photons have:
[tex]2.00 mole= 2.00\times 6.022\times 10^{23} =1.2044\times 10^{24} [/tex] photons
Total energy of the [tex]1.0244\times 10^{24} [/tex] photons = E'
[tex]E'=1.3623\times 10^{-19} J\times 1.0244\times 10^{24}=164,075.41 J=164.08 kJ[/tex]
The energy of 2.00 mol of photons of an infrared radiation 1460 nm is 164.08 kilo Joules.
