Question

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?

Answer 1

E= 10.2 ev

`Wavelength= 1.21 * 10^(-7) \ m`

`Frequency =2.479 * 10^(15) \ s^(-1)`

Step-by-step explanation:

It is given that energy in first exited state (n=2) ,
`E_2`= -3.4 ev

Also , it is given than energy in ground state (n=1) ,
`E_1`= -13.6 ev.

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

Energy of photon released,
`E=E_(final)-E_(initial)`

Therefore,
`E=(-3.4) - (-13.6)\ ev=10.2\ ev`.

To convert energy from ev(electron volt) into joule we need to multiply energy in ev by charge of electron which is (
`1.6 * 10^(-19)` )

Therefore,
`E_(joule) = E_(ev)* 1.6 * 10^(-19) \ joules`

Now , we know that photon is an quantum of light . Therefore, speed of photon in vaccum is equal to speed of light which is ,
`c=3* 10^8\ m/s`.

Now, by energy-wavelength relation,

`E_(joule)=(h* c)/(\lambda)`

where ,
`h=6.626* 10^(-34)\ joule-second`

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

`10.2 * 1.6* 10^(-19)=(6.626*10^(-34) * 3 * 10^8)/(\lambda)`

`\lambda=\frac{{6.626*10^(-34) * 3 * 10^8}}{10.2 * 1.6* 10^(-19)}`

`\lambda=1.21 * 10^(-7)\ m`

We know,
`c=\lambda* frequency`

Therefore,
`frequency=(c)/(\lambda)= (3* 10^8)/(1.21* 10^-7) \ s^(-1)=2.479* 10^(15) \ s^(-1)`

Hence, this is the required solution.