Question

An atom in its ground state is excited when it absorbs a single photon of light. The atom then relaxes back to the ground state by emitting two photons, the first, a red photon at 700 nm, and the second, an infrared photon at 1750 nm. What is the wavelength of the absorbed photon? 500 nm 1225 nm 700 nm 1950 nm 1750 nm

Answer 1

Answer: Option (a)  is the correct answer.

Explanation:

The given data is as follows.

         wavelength of red photon ($\lambda_{1}$) = 700 nm

         wavelength of infrared photon ($\lambda_{2}$) = 1750 nm

Therefore, calculate the wavelength of absorbed photon as follows.

    $\frac{1}{\lambda_{abs}} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$

              = $\frac{1}{700} + \frac{1}{1750}$

              = $\frac{2450}{1225000}$

or,            $\lambda_{abs}$ = $\frac{1225000}{2450}$              

                            = 500 nm

Therefore, we can conclude that the wavelength of the absorbed photon is 500 nm.

Answer 2

Answer:

500 nm

Explanation:

We are given that

Wavelength of red photon=$\lambda_1=700$nm

Wavelength of Infrared photon=$\lambda_2=1750$nm

We have to find the length of absorbed photon.

We know that

Energy of photon=$\frac{hc}{\lambda}$

Where $\lambda$=Wavelength of photon

Energy of absorbed photon=Sum of energy of emitting two photons.

$\frac{hc}{\lambda}=\frac{hc}{\lambda_1}+\frac{hc}{\lambda_2}$

$\frac{hc}{\lambda}=hc(\frac{1}{\lambda_1}+\frac{1}{\lambda_2})$

$\frac{1}{\lambda}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$

Substitute the values then we get

$\frac{1}{\lambda}=\frac{1}{700}+\frac{1}{1750}=\frac{1750+700}{1225000}=\frac{2450}{1225000}$

$\lambda=\frac{1225000}{2450}=500$

Hence, the wavelength of absorbed photon=500 nm

Option A is true.