Question

Calculate the change in energy of an atom that emits a photon of wavelength 2.21 meters. (Planck's constant is 6.626 x 10-34 joule seconds, the speed of light is 2.998 x 108 m/s) A. 8.9886 x10-26 joules B. 4.8844 x 10-42 joules second C. 1.9864 x 10-25 joules D. 1.4643 x 10-33 joules /second

Answer 1

9.01 × 10⁻²⁶ J

Explanation

ΔE = h · f

Where

• ΔE the change in energy,
• h the planck's constant, and
• f frequency of the emission.

However, only λ is given.

f = c / λ

Where

• f frequency of the emission,
• λ wavelength of the emission, and
• c the speed of light.

For this emission:

f = 2.998 × 10⁸ / 2.21 = 1.36 × 10⁸ s⁻¹.

ΔE = h · f = 6.626 × 10⁻³⁴ × 1.36 × 10⁸ = 9.01 × 10⁻²⁶ J