An electron in a hydrogen atom in the level n = 5 undergoes a transition to level n = 2. What does the Bohr model of the atom predict for the frequency of the emitted radiation?

Answer 1

1) We can first use the Rydberg equation to find the energy of the emitted radiation

E = -R_H × [ (1/n_i²) - (1/n_f²) ]

where R_H is the Rydberg constant of 2.18 x 10^-18 J/electron, n_i is the initial energy level, and n_f is the final energy level

2) Let's identify the initial and final energy levels

The initial energy level, n_i, is equal to 5 since the electron was originally in the n=5 energy level.

The final energy level, n_f, is equal to 2 since the electron ends up in the n=2 energy level.

3) We are now ready to solve for the energy of the emitted radiation

E = -R_H × [ (1/n_i²) - (1/n_f²) ]

E = -2.18 x 10^-18 J/electron × [ (1/5²) - (1/2²) ]

E = -2.18 x 10^-18 J/electron × [ (1/25) - (1/4) ] (The common denominator is 100. We can therefore convert convert 1/25 to 4/100 and 1/4 to 25/100.)

E = -2.18 x 10^-18 J/electron × [ (4/100) - (25/100) ]

E = -2.18 x 10^-18 J/electron × (-21/100) (Instead of diving by 100 we can instead multiply by 10^-2)

E = -2.18 x 10^-18 J/electron × -21 × 10^-2

E = 45.78 x 10^-20 (We need to multiply 45.78 by 10^-1 to move the decimal over to the left one place. We then have to also multiply by 10¹ so we aren't actually changing anything.)

E = 4.578 x 10^-19 J

3) We can then use the equation that relates energy to frequency to solve for frequency

E = h x f

where E is energy, h is Planck's constant of 6.626 x 10^-34 J/s, and f is frequency

f = E / h

f = (4.578 x 10^-19 J) / 6.626 x 10^-34 J/s (Dividing by 10^-34 is the same as multiplying by 10³⁴)

f = 0.6909 x 10¹⁵ 1/s (To move the decimal to the right one, we will multiplying by 10¹. We then need to multiply by 10^-1 so we aren't actually changing anything.)

f = 6.909 x 10¹⁴ Hz (We use the units of Hz since Hz is equal to 1/s)

4) The frequency of emitted radiation is therefore 6.909 x 10¹⁴ Hz