Answer :
Answer : The energy of one photon of hydrogen atom is, [tex]3.03\times 10^{-19}J[/tex]
Explanation :
First we have to calculate the wavelength of hydrogen atom.
Using Rydberg's Equation:
[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 = 10973731.6 m⁻¹
[tex]n_f[/tex] = Higher energy level = 3
[tex]n_i[/tex]= Lower energy level = 2
Putting the values, in above equation, we get:
[tex]\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\lambda=6.56\times 10^{-7}m[/tex]
Now we have to calculate the energy.
[tex]E=\frac{hc}{\lambda}[/tex]
where,
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = [tex]6.56\times 10^{-7}m[/tex]
Putting the values, in this formula, we get:
[tex]E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}[/tex]
[tex]E=3.03\times 10^{-19}J[/tex]
Therefore, the energy of one photon of hydrogen atom is, [tex]3.03\times 10^{-19}J[/tex]