In a ruby laser, an electron jumps from a higher energy level to a lower one. if the energy difference between the two levels is 1. 8 ev, what is the wavelength of the emitted photon?

Respuesta :

The wavelength of the emitted photon is([tex]\lambda[/tex])= 690nm

How can we calculate the wavelength of the emitted photon?

To calculate the wavelength of the photon we are using the formula,

[tex]\triangle E= \frac{h\times c}{\lambda}[/tex]

Or,[tex]\lambda= \frac{h\times c}{\triangle E}[/tex]

We are given here,

[tex]\triangle E[/tex]= The energy difference between the two levels = 1. 8 ev= [tex]1.8\times 1.6 \times 10^{-19}[/tex] C.

h= Planck constant = [tex]6.626\times 10^{-34}[/tex] Js.

c= speed of light = [tex]3\times10^8[/tex] m/s.

We have to find the wavelength of the emitted photon =[tex]\lambda[/tex] m.

Therefore, we substitute the known parameters in the above equation, we can find that,

[tex]\lambda= \frac{h\times c}{\triangle E}[/tex]

Or,[tex]\lambda= \frac{6.626\times 10^{-34}\times 3\times 10^8}{1.8\times 1.6 \times 10^{-19}}[/tex]

Or,[tex]\lambda= 690\times 10^{-9}[/tex] m

Or,[tex]\lambda[/tex]=690 nm.

From the above calculation we can conclude that the wavelength of the emitted photon is 690nm.

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