Q.

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)

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

answer is 50.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Energy emitted in 0.1 sec. = 0.1 sec. × 10−3J/s = 10−4 JIf 'n' photons of λ = 1000 nm are emitted, then 10−4 =  nhcλ⇒10−4=n×6.63×10−34×3×1081000×10−9⇒ n = 5.02 × 1014 = 50.2 × 1013⇒ 50 (nearest integer)
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon