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As mentioned in this week’s notes on page 4, the electrons of an atom can occupy different energy shells within the atom (similar to how the planets all occupy different orbits around the Sun). Electrons prefer to be in the lowest energy shell possible (the ground state); however, they can gain energy and jump to a higher shell by absorbing light or being excited by an electric current. In accordance with the conservation of energy, if an electron drops from a higher energy level to a lower one, this must emit a photon (particle of light) with energy equal to the energy difference of the shells. A Balmer series transition is any transition of an electron from some higher energy shell down to the second lowest energy shell (n=2) in hydrogen. Looking at image (b) above, what is the wavelength of a photon emitted during the Balmer transition from the n=3 shell in hydrogen? (remember nm is short for a nanometer, for example 656 nm = 656 x 10-9 meters)
656E-9 meters

486E-9 meters

434E-9 meters

410E-9 meters

Respuesta :

Edufirst
Answer: 656 E -9 meters

Explanation:

The figure that accompanies the question shows the wavelenghts of the photons emitted according to Balmer series transition , from energy levels (n) 3, 4, 5, and 6 to the energy level (n) 2, in  hydrogen atoms.

These are the values shown in the figure

Transition                     wavelength of the photon emitted
                                                 nm

from n = 3 to n = 2                  656 <------------- this is the value requested

from n = 4 to n = 2                  486

from n = 5 to n= 2                   434

from n = 6 to n = 2                  410

The wavelength of a photon emitted from the n = 3 shell in hydrogen is the first data of the table, i.e 656 nm.

Using the conversion factor from nm to m that result is:

656 nm * 1 m / (10^9 nm) =  656 * 10 ^ - 9 m.

Answer: The wavelength of the photon will be [tex]656\times 10^{-9}m[/tex]

Explanation: Using Rydberg's Equation:

[tex]\frac{1}{\lambda }=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right)[/tex]

Where,  [tex]\lambda=\text{Wavelength of radiation}=?[/tex]

[tex]R_H=\text{Rydberg's Constant}=1.09737\times 10^7m^{-1}[/tex]

[tex]n_i[/tex] = Initial energy level =  3

[tex]n_f[/tex]=final energy level = 2  (Balmer series)

Putting the values, in above equation, we get

[tex]\frac{1}{\lambda }=1.09737\times 10^7m^{-1\left(\frac{1}{2^2}-\frac{1}{3^2} \right)[/tex]

[tex]{\lambda }= 6.56\times 10^{-7}m=656\times 10^{-9}m[/tex]






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