Question

An electron transition in the hydrogen atom emits a photon with a wavelength of 1093.5 nm. calculate ∆e for this transition in aj. include the correct sign for the emission of a photon.

Answer 1

To calculate ΔE (change in energy) for the electron transition in the hydrogen atom, use the formula ΔE = hc/λ where h is Planck's constant (6.626×10-34 J·s), c is the speed of light (2.998 × 108 m/s), and λ is the wavelength. Substituting the values, the change in energy is approximately 1.817 × 10-19 J, with the emission of the photon signifying the release of energy.

Step-by-step explanation:

To calculate ΔE (change in energy) for the electron transition in the hydrogen atom, we can use the formula:

ΔE = hc/λ

Where ΔE is the change in energy, h is Planck's constant (6.626×10-34 J·s), c is the speed of light (2.998 × 108 m/s), and λ is the wavelength (1093.5 nm = 1093.5 × 10-9 m).

Substituting the values into the formula:

ΔE = (6.626×10-34 J·s)(2.998 × 108 m/s) / (1093.5 × 10-9 m)

ΔE ≈ 1.817 × 10-19 J

The change in energy (ΔE) for this electron transition is approximately 1.817 × 10-19 J, with the emission of the photon signifying the release of energy.