Answer:
E= 10.2 ev
[tex] Wavelength= 1.21 \times 10^{-7} \ m[/tex]
[tex]Frequency =2.479 \times 10^{15} \ s^{-1}[/tex]
Explanation:
It is given that energy in first exited state (n=2) , [tex]E_2[/tex]= -3.4 ev
Also , it is given than energy in ground state (n=1) , [tex]E_1[/tex]= -13.6 ev.
We know energy of photon released is difference of the final and initial level of electron.
Energy of photon released, [tex]E=E_{final}-E_{initial}[/tex]
Therefore, [tex]E=(-3.4) - (-13.6)\ ev=10.2\ ev[/tex].
To convert energy from ev(electron volt) into joule we need to multiply energy in ev by charge of electron which is ( [tex]1.6 \times 10^{-19}[/tex] )
Therefore, [tex]E_{joule} = E_{ev}\times 1.6 \times 10^{-19} \ joules[/tex]
Now , we know that photon is an quantum of light . Therefore, speed of photon in vaccum is equal to speed of light which is , [tex]c=3\times 10^8\ m/s[/tex].
Now, by energy-wavelength relation,
[tex]E_{joule}=\dfrac{h\times c}{\lambda}[/tex]
where , [tex]h=6.626\times 10^{-34}\ joule-second[/tex]
Now, putting values of h ,c and E in above equation .
[tex]10.2 \times 1.6\times 10^{-19}=\dfrac{6.626\times10^{-34} \times 3 \times 10^8}{\lambda}[/tex]
[tex]\lambda=\dfrac{{6.626\times10^{-34} \times 3 \times 10^8}}{10.2 \times 1.6\times 10^{-19}}[/tex]
[tex]\lambda=1.21 \times 10^{-7}\ m[/tex]
We know, [tex]c=\lambda\times frequency[/tex]
Therefore, [tex]frequency=\dfrac{c}{\lambda}= \dfrac{3\times 10^8}{1.21\times 10^-7} \ s^{-1}=2.479\times 10^{15} \ s^{-1}[/tex]
Hence, this is the required solution.
