Question:

An electron transitions from the fourth energy level (n=4) to the second energy level (n=2) within a hydrogen atom. Determine:

a) The energy change (ΔE) in electron volts (eV) associated with this transition.

b) The wavelength (λ) of light emitted in this transition.

Given:

• The energy of an electron in an energy level (E) is given by the formula E = -13.6 eV / n², where n is the principal quantum number.
• The speed of light (c) is approximately 3.00 × 10⁸ m/s.
• The Planck's constant (h) is approximately 6.63 × 10⁻³⁴ J·s.

Answer:

a) The energy change (ΔE) in electron volts (eV) associated with this transition can be calculated using the formula:

ΔE = (E_final - E_initial)

Where E_final is the energy of the final level and E_initial is the energy of the initial level.

Given that the initial energy level (n=4) and final energy level (n=2), we can substitute these values into the formula:

E_initial = -13.6 eV / (4²) = -13.6 eV / 16 = -0.85 eV E_final = -13.6 eV / (2²) = -13.6 eV / 4 = -3.4 eV

Now, we can calculate the energy change (ΔE):

ΔE = (-3.4 eV) - (-0.85 eV) = -2.55 eV

Therefore, the energy change associated with this transition is ΔE = -2.55 eV.

b) The wavelength (λ) of light emitted in this transition can be determined using the equation:

ΔE = hc / λ

Where ΔE is the energy change in joules (J), h is Planck's constant, c is the speed of light, and λ is the wavelength.

To convert the energy change from electron volts (eV) to joules (J), we will use the conversion factor:

1 eV = 1.6 × 10⁻¹⁹ J

Converting the energy change from eV to J:

ΔE = -2.55 eV × (1.6 × 10⁻¹⁹ J/eV) = -4.08 × 10⁻¹⁹ J

Now, we can rearrange the equation to solve for the wavelength (λ):

λ = hc / ΔE

Substituting the values:

λ = (6.63 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / (-4.08 × 10⁻¹⁹ J) = 1.54 × 10⁻⁷ m

Therefore, the wavelength of light emitted in this transition is λ = 1.54 × 10⁻⁷ m, or approximately 154 nanometers (nm).