The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x × 10¹³. The value of x is _____________. (Nearest integer)
(h = 6.63 × 10⁻³⁴ Js, c = 3.00 × 10⁸ ms⁻¹)

Solution

Chemistry ma’am · 3 steps

Find the total energy the beam gives out, then divide by the energy of one photon.

1. Total energy. Power is energy per second. So in 0.1 s the range finder emits E = Pt = (1 × 10⁻³ W) × (0.1 s) = 1 × 10⁻⁴ J.
2. One photon. Convert the wavelength to metres: 1000 nm = 1 × 10⁻⁶ m. Then Eₚₕₒₜₒₙ = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(1 × 10⁻⁶) = 1.989 × 10⁻¹⁹ J.

   N = Pt/(hc/λ) = (1 × 10⁻⁴)/(1.989 × 10⁻¹⁹) = 5.03 × 10¹⁴

3. So. N = 50.3 × 10¹³, so to the nearest integer x = 50.

The common slip: 503. Leaving out the time and dividing the power alone by hc/λ gives the photons per second, 5.03 × 10¹⁵, so x comes out as 503 instead of 50.

The idea to keep. Number of photons = total energy ÷ energy of one photon, with total energy = Pt. Convert mW to W and nm to m before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.