Question

What is the wavelength λ of the photon that has been released in Part B? ΔE = −2.04×10−18 J The energy of the photon emitted by the electron is related to its wavelength by E=hcλ where λ is the wavelength in meters, 6.626×10−34 J⋅s is Planck's constant, and 3.00×108 m/s is the speed of light in a vacuum.

Answer 1

The wavelength λ of the photon that has been released in Part B is approximately 9.67 × 10^-7 meters.

Step-by-step explanation:

The energy of a photon emitted by an electron is related to its wavelength by the equation E=hc/λ, where E is the energy in joules, λ is the wavelength in meters, h is Planck's constant (6.626×10^-34 J⋅s), and c is the speed of light in a vacuum (3.00×10^8 m/s).

To find the wavelength, we can rearrange the equation as λ = hc/E. Plugging in the given value for ΔE (-2.04×10^-18 J) into the equation, we get λ = (6.626×10^-34 J⋅s * 3.00×10^8 m/s) / (-2.04×10^-18 J).

Simplifying the equation gives us the wavelength λ ≈ 9.67 × 10^-7 meters.