Question:

An electron in a hydrogen atom undergoes a transition from the n=3 energy level to the n=2 energy level. During this transition, the electron emits a photon with a wavelength of 486 nm.

1. Calculate the frequency (in Hz) of this emitted photon.

2. Determine the energy (in electron volts) associated with this transition.

3. Given that the energy of an electron in the nth energy level of a hydrogen atom is given by the equation:

   E = -13.6/n^2

   where E is the energy in electron volts and n is the principal quantum number, calculate the energy change (in electron volts) for this transition.

Assume Planck's constant, h, is 6.626 x 10^-34 J·s and 1 eV = 1.6 x 10^-19 J.

Answer:

1. The frequency (f) of a photon can be determined using the formula:

   c = λf

   where c is the speed of light (3 x 10^8 m/s) and λ is the wavelength of the photon.

   Given λ = 486 nm = 486 x 10^-9 m, we can solve for f:

   f = c / λ

   f = (3 x 10^8 m/s) / (486 x 10^-9 m)

   f = 6.17 x 10^14 Hz

   Therefore, the frequency of the emitted photon is 6.17 x 10^14 Hz.

2. The energy (E) associated with a photon can be determined using the equation:

   E = hf

   where h is Planck's constant (6.626 x 10^-34 J·s) and f is the frequency of the photon.

   Given f = 6.17 x 10^14 Hz, we can solve for E:

   E = (6.626 x 10^-34 J·s) * (6.17 x 10^14 Hz)

   E = 4.08 x 10^-19 J

   Since 1 eV = 1.6 x 10^-19 J, we can convert the energy to electron volts:

   E = (4.08 x 10^-19 J) / (1.6 x 10^-19 J/eV)

   E = 2.55 eV

   Therefore, the energy associated with this transition is 2.55 electron volts.

3. To calculate the energy change (ΔE) for this transition, we subtract the initial energy (Ei) from the final energy (Ef):

   ΔE = Ef - Ei

   Given Ei = -13.6 / 3^2 = -13.6 / 9 eV and Ef = -13.6 / 2^2 = -13.6 / 4 eV, we can calculate ΔE:

   ΔE = (-13.6 / 4 eV) - (-13.6 / 9 eV)

   ΔE = (-3.4 eV) - (-1.51 eV)

   ΔE = -1.89 eV

   Therefore, the energy change for this transition is -1.89 electron volts.