This is an incomplete question, the image for the given question is attached below.
Answer : The wavelength of photon would be absorbed, [tex]3.06\times 10^{-7}m[/tex]
Explanation :
From the given diagram of energy we conclude that,
Energy at ground state, A = 400 zJ
Energy of 2nd excited state, C = 1050 zJ
Now we have to calculate the energy of the photon.
[tex]E=E_c-E_A[/tex]
[tex]E=(1050-400)zJ= 650zJ=650\times 10^{-21}J[/tex]
Now we have to calculate the wavelength of the photon.
Formula used :
[tex]E=h\times \nu[/tex]
As, [tex]\nu=\frac{c}{\lambda}[/tex]
So, [tex]E=h\times \frac{c}{\lambda}[/tex]
where,
E = energy of photon = [tex]650\times 10^{-21}J[/tex]
[tex]\nu[/tex] = frequency of photon
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\lambda[/tex] = wavelength of photon = ?
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get:
[tex]650\times 10^{-21}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}[/tex]
[tex]\lambda=3.06\times 10^{-7}m[/tex]
Therefore, the wavelength of photon would be absorbed, [tex]3.06\times 10^{-7}m[/tex]
