contestada

a modern compact fluorescent lamp contains 1.4 mg of mercury (hg). if each mercury atom in the lamp were to emit a single photon of wavelength 508 nm, how many joules of energy would be emitted?

Respuesta :

The fluorescent lamp emits 1.64 Joules of energy.

The energy can be calculated using the formula -

E = Planck's constant × speed of light ÷ wavelength

The value of Planck's constant is 6.6×[tex] {10}^{ - 34} [/tex] J/s. Speed of light is 3×10⁸ m/s. Wavelength = 508×[tex] {10}^{ - 9} [/tex] meters

Keep the values in formula -

Energy = (6.6×[tex] {10}^{ - 34} [/tex] × 3×10⁸) ÷ 508×[tex] {10}^{ - 9} [/tex]

Energy = 3.9×[tex] {10}^{ - 19} [/tex] Joules

Number of moles = mass ÷ molar mass

Number of moles = 0.0014 ÷ 200.6

Performing division

Number of moles = 7×[tex] {10}^{ - 6} [/tex]

Number of atoms = number of moles × Avogadro's number

Number of atoms = 7×[tex] {10}^{ - 6} [/tex] × 6.022×10²³

Performing multiplication

Number of atoms = 4.2×10¹⁸

Total energy emitted = 4.2×10¹⁸×3.9×[tex] {10}^{ - 19} [/tex]

Performing multiplication

Total energy emitted = 1.64 Joules

Therefore, energy emitted is 1.64 Joules.

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