Respuesta :
Answer:
The longest wavelength observed in the Balmer series of the H atom spectrum is 656.3 nm.
The shortest wavelength observed in the Balmer series of the H atom spectrum is 364.6 nm.
Explanation:
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
[tex]n_f[/tex] = Higher energy level
[tex]n_i[/tex]= Lower energy level
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
For wavelength to be longest, energy would be minimum, i.e the electron will jump from third level to second level :
[tex]n_f[/tex] = Higher energy level = [tex]3[/tex]
[tex]n_i[/tex]= Lower energy level = 2 (Balmer series)
Putting the values, in above equation, we get
[tex]\frac{1}{\lambda_{Balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\lambda_{Balmer}=\frac{36}{5R_H}[/tex]
[tex]\lambda_{Balmer}=\frac{36}{5\times 1.097\times 10^7 m^-1}}=6.563\times 10^{-7} m=656.3 nm[/tex]
[tex]1 m =10^9 nm[/tex]
The longest wavelength observed in the Balmer series of the H atom spectrum is 656.3 nm.
For wavelength to be shortest, energy would be maximum, i.e the electron will from infinite level to second level. :
[tex]n_f[/tex] = Higher energy level = [tex]\infty[/tex]
[tex]n_i[/tex]= Lower energy level = 2 (Balmer series)
Putting the values, in above equation, we get
[tex]\frac{1}{\lambda_{Balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{\infty^2} \right )[/tex]
[tex]\lambda_{Balmer}=\frac{4}{R_H}[/tex]
[tex]=\frac{4}{1.097\times 10^7 m^-1}=3.646\times 10^{-7} m=364.6 nm[/tex]
The shortest wavelength observed in the Balmer series of the H atom spectrum is 364.6 nm.
