Question

A source produces monochromatic light of frequency 5.0 × 10¹⁴ Hz and the power emitted is 3.31 mW. The number of photons emitted per second by the source, on an average is ______.

पर्याय

• 10¹⁶

• 10²⁴

• 10¹⁰

• 10²⁰

Answer 1

A source produces monochromatic light of frequency 5.0 × 10¹⁴ Hz and the power emitted is 3.31 mW. The number of photons emitted per second by the source, on an average is 10¹⁶.

Explanation:

Given: v = 5.0 × 10¹⁴ Hz

E = hv

= 6.63 × 10⁻³⁴ × 5 ×10¹⁴

= 33.15 × 10⁻²⁰ J

Now,

n = `P/E`

= `(3.31 xx 10^-3)/(33.15 xx 10^-20)`

= `11/10 xx 10^17`

≈ 10¹⁶