The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (given h=6×10⁻³⁴ J-s)

Options

(a) 100
(b) 1000
(c) 3×10²⁰
(d) 3×10¹⁸

Correct Answer:

3×10²⁰

Explanation:

Power = nhc / ( × t)

⇒ 100 = [n × (6 × 10⁻³⁴) × (3 × 10⁸)] / (540 × 10⁹ × 1)

⇒ n × 18 × 10⁻²⁶ / 540 × 10⁻⁹ = 100

⇒ n = 100 × 540 × 10⁻⁹ / 18 × 10⁻²⁶ = 3 x 10²⁰