Question

Calculate the wavelength of the photon that would be absorbed or emitted. Round your answer to 3 significant digits.

Answer 1

This is an incomplete question, the image for the given question is attached below.

Answer : The wavelength of photon would be absorbed, $3.06\times 10^{-7}m$

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.

$E=E_c-E_A$

$E=(1050-400)zJ= 650zJ=650\times 10^{-21}J$

Now we have to calculate the wavelength of the photon.

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

E = energy of photon = $650\times 10^{-21}J$

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon  = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$650\times 10^{-21}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=3.06\times 10^{-7}m$

Therefore, the wavelength of photon would be absorbed, $3.06\times 10^{-7}m$