Question

An electric bulb of 100 w converts 3% of electrical energy into light energy. If the wavelength of light emitted is 6625 Å, the number of photons emitted is 1 s is……..

(h = 6.625 × 10^–34 J.s)

(A) 10¹⁷ 

(B) 10¹⁹

(C) 10²¹

(D) 10¹⁵

Answer 1

The correct option is (B) 10¹⁹.

Explanation:

3% electric energy converted to light energy 

Hence power emitted = 100 × {3 / (100)} = 3w 

Also power = {(nhc) / λ} 

∴ n = {Pλ / hc} 

= [{3 × 6625 × 10^–10} / {6.62 × 10^–34 × 3 × 108}] 

= 1000 × 10¹⁶ 

= 10¹⁹ per second. 

i.e. no of photons emitted per second is 10¹⁹