Question

A yellow line occurs at 572.8 nm in the emission spectrum of phosphorus. What amount of energy, in joules, must be released by an electron in a phosphorus atom to produce a photon of this light?

Answer 1

3.47 × 10⁻¹⁹ J

Step-by-step explanation:

A yellow line occurs at 572.8 nm in the emission spectrum of phosphorus. When an electron undergoes this process, it requires an amount of energy to produce a photon of this light. We can calculate the energy (E) using the following expression.

E = h × c/λ

where,

• h: Planck's constant (6.63 × 10⁻³⁴ J.s)

• c: speed of light (3.00 × 10⁸ m/s)

• λ: wavelength (572.8 nm = 5.728 × 10⁻⁷ m)

E = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s/5.728 × 10⁻⁷ m

E = 3.47 × 10⁻¹⁹ J