The energy (in joules) of an electron energy level in the Bohr atom is given by the expression: En = -2.179E-18/n2 J where n is the quantum number that specifies the energy level. What is the frequency in Hz of the radiation emitted when an electron falls from level n = 3 to level n = 2?
n1 = 2; n1^2 = 4
n2 = 3; n2^2 = 9
E = 2.179E-18(1/4-1/9)
Solve for E, then E = hf and solve for f.
To find the frequency of the radiation emitted when an electron falls from level n = 3 to level n = 2, we need to use the equation that relates the energy of a photon (radiation) to its frequency:
E = h*f
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the radiation.
First, we need to find the energy difference between the two energy levels:
ΔE = E_final - E_initial = (-2.179 x 10^-18 J / (3^2)) - (-2.179 x 10^-18 J / (2^2))
ΔE = (-2.179 x 10^-18 J / 9) - (-2.179 x 10^-18 J / 4)
ΔE = -2.42 x 10^-19 J
Now, we have the energy difference, so we can calculate the frequency of the radiation using the equation:
ΔE = h*f
f = ΔE / h
f = (-2.42 x 10^-19 J) / (6.626 x 10^-34 J·s)
f ≈ 3.66 x 10^14 Hz
Therefore, the frequency of the radiation emitted when an electron falls from level n = 3 to level n = 2 is approximately 3.66 x 10^14 Hz.