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

Answer:    

E= 10.2 ev

$Wavelength= 1.21 \times 10^{-7} \ m$

$Frequency =2.479 \times 10^{15} \ s^{-1}$

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 ( $1.6 \times 10^{-19}$ )

Therefore, $E_{joule} = E_{ev}\times 1.6 \times 10^{-19} \ joules$

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

Now, by energy-wavelength relation,

$E_{joule}=\dfrac{h\times c}{\lambda}$

where , $h=6.626\times 10^{-34}\ joule-second$

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

$10.2 \times 1.6\times 10^{-19}=\dfrac{6.626\times10^{-34} \times 3 \times 10^8}{\lambda}$

$\lambda=\dfrac{{6.626\times10^{-34} \times 3 \times 10^8}}{10.2 \times 1.6\times 10^{-19}}$

$\lambda=1.21 \times 10^{-7}\ m$

We know, $c=\lambda\times frequency$

Therefore, $frequency=\dfrac{c}{\lambda}= \dfrac{3\times 10^8}{1.21\times 10^-7} \ s^{-1}=2.479\times 10^{15} \ s^{-1}$

Hence, this is the required solution.