What is the energy of a photon released by an electron going from the first excited energy (n=2) level to ground state (n=1). note: e1 = -13.6ev. e2 = -3.4ev. what is the wavelength, frequency, and speed of the photon in a vacuum?

E= 10.2 ev

Explanation:

It is given that energy in first exited state (n=2) , = -3.4 ev

Also , it is given than energy in ground state (n=1) , = -13.6 ev.

We know energy of photon released is difference of the final and initial level of electron.

Energy of photon released,  

Therefore, .

To convert energy from ev(electron volt) into joule we need to multiply energy in ev by charge of electron which is ( )

Therefore,

Now , we know that photon is an quantum of light . Therefore, speed of photon in vaccum is equal to speed of light which is , .

Now, by energy-wavelength relation,

where ,

Now, putting values of h ,c and E in above equation .

We know,

Therefore,

Hence, this is the required solution.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14