A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x × 10²⁰. The value of x is __________. [Given : h = 6.63 × 10⁻³⁴ Js and c = 3.0 × 10⁸ ms⁻¹]

Solution

Chemistry ma’am · 3 steps

Power is energy per second, so divide the bulb's 50 joules each second by the energy of one photon.

1. Energy of one photon. Each photon carries E = hc/λ. Convert the wavelength first: 795 nm = 7.95 × 10⁻⁷ m.

   E = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(7.95 × 10⁻⁷) = (1.989 × 10⁻²⁵)/(7.95 × 10⁻⁷) = 2.502 × 10⁻¹⁹ J

2. Count the photons. A 50 W bulb gives out 50 J every second. The number of photons per second is N = 50/(2.502 × 10⁻¹⁹) = 1.998 × 10²⁰.
3. So. N ≈ 2 × 10²⁰ photons per second, so x = 2.

The common slip: 0.2. Taking 1 nm as 10⁻¹⁰ m makes each photon 10 times more energetic and gives 0.2, not 2. Remember 1 nm is 10⁻⁹ m.

The idea to keep. Photons per second = power ÷ energy of one photon, with E = hc/λ. Put λ in metres before you divide.

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