Question

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?

Answer 1

The wavelength of the emitted photon is($\lambda$)= 690nm

How can we calculate the wavelength of the emitted photon?

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

$\triangle E= \frac{h\times c}{\lambda}$

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

We are given here,

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

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

c= speed of light = $3\times10^8$ m/s.

We have to find the wavelength of the emitted photon =$\lambda$ m.

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

$\lambda= \frac{h\times c}{\triangle E}$

Or,$\lambda= \frac{6.626\times 10^{-34}\times 3\times 10^8}{1.8\times 1.6 \times 10^{-19}}$

Or,$\lambda= 690\times 10^{-9}$ m

Or,$\lambda$=690 nm.

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

Learn more about ruby laser:

brainly.com/question/17245697

#SPJ4