Answer: [tex]1.8\times 10^{-19}J[/tex]
Explanation:
Using Rydberg's Equation for hydrogen atom:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level
[tex]n_i[/tex]= Lower energy level
We have:
[tex]n_f=6, n_i=3[/tex]
[tex]R_H=1.09\times 10^7 m^{-1}[/tex]
[tex]\frac{1}{\lambda}=1.09\times 10^7 m^{-1}\times \left(\frac{1}{3^2}-\frac{1}{6^2} \right )[/tex]
[tex]\frac{1}{\lambda}=1.09\times 10^7 m^{-1}\times \frac{3}{36}[/tex]
[tex]\frac{1}{\lambda}=0.0908\times 10^{7}m^{-1}[/tex]
[tex]\lambda=11.0\times 10^{-7} m[/tex]
The relation between energy and wavelength of light is given by Planck's equation, which is:
[tex]E=\frac{hc}{\lambda}[/tex]
where,
E = energy of the light
h = Planck's constant
c = speed of light
[tex]\lambda[/tex] = wavelength of light
[tex]E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{11.0\times 10^{-7}m}[/tex]
[tex]E=1.8\times 10^{-19}J[/tex]
Thus energy of a photon that is ejected from a hydrogen atom when an electron relaxes from the 6th to the 3rd energy level is [tex]1.8\times 10^{-19}J}[/tex]