Respuesta :
Answer: [tex]0.31\times 10^{16}Hz[/tex].
Explanation:
[tex]E=\frac{hc}{\lambda}[/tex]
[tex]\lambda[/tex] = Wavelength of radiation
E= energy
Using Rydberg's Equation:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation = ?
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level = 4
[tex]n_i[/tex]= Lower energy level = 1
Z= atomic number = 1 (for hydrogen)
Putting the values, in above equation, we get
[tex]\frac{1}{\lambda}=10973731.6m^{-1}\left(\frac{1}{1^2}-\frac{1}{4^2} \right )\times 1[/tex]
[tex]\lambda=9.7\times 10^{-8}m[/tex]
The relationship between wavelength and frequency of the wave follows the equation:
[tex]\nu=\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of the wave = ?
c = speed of light = [tex]3\times 10^8ms^{-1}[/tex]
[tex]\lambda [/tex] = wavelength of the wave = [tex]9.7\times 10^{-8}m[/tex]
[tex]\nu=\frac{3\times 10^8ms^{-1}}{9.7\times 10^{-8}m}[/tex]
[tex]\nu=0.31\times 10^{16}s^{-1}=0.31\times 10^{16}Hz[/tex]
The frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is [tex]0.31\times 10^{16}Hz[/tex].
