Question

When 50 electrons drop from energy level 4 to energy level 2, how many photons of light are given off?

1 photon

100 photons

2 photons

50 photons

Answer 1

The number of photons given off will be 50 photons.

To find the answer, we need to know about the plank's equation.

How to find the number of photons emitted?

• We have the expression for energy of a single electron in eV as,

                     $E=-13.6(\frac{1}{(n_f)^2}- \frac{1}{(n_i)^2})eV$

• We have,

                     $n_f=4\\n_i=2\\N=50$

• Substituting values, we get,

                  $E=-13.6(\frac{1}{(4)^2}- \frac{1}{(2)^2})eV\\\\E=-13.6*-0.188=2.55eV$

• For N electrons,

                $E=50*2.55eV=127.5eV$

• We have the plank's equation,

                            E=nhf

• From this, the number of photons emitted from 50 electrons will be,

                    $n=\frac{E}{h*f} =\frac{127.5*1.67*10^{-19}J}{(6.63*10^-34)Js}$

• To find n, we have to find the frequency f. For that, we have the equation,

                       $\frac{1}{wave length}=R_H(\frac{1}{(n_i)^2}- \frac{1}{(n_f)^2})\\\\1/wv= 1.1*10^5(\frac{1}{(2)^2}- \frac{1}{(4)^2})=20625cm^{-1}.\\wavelength=486nm.$

• Thus, frequency will be,

                 $f=\frac{c}{wavelength} =\frac{3*10^8}{486*10^{-9}} =6.172*10^{14}s{-1}$

• Then, the number of photons will be,

                       $n=\frac{E}{h*f} =\frac{127.5*1.67*10^{-19}J}{(6.63*10^-34)Js*6.17*10^{14} s^{-1}} \\\\n=52.05 photons$

Thus, we can conclude that, the number of photons given off will be 50 photons.

Learn more about the photons here:

https://brainly.com/question/15946945

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