Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
17.3k views
in Chemistry by (64.7k points)

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second.

1 Answer

+1 vote
by (62.3k points)
selected by
 
Best answer

Power of the bulb = 25 W = 25 Js-1

Wave length of light λ = 0.57 μm = 0.57 x 10-6 m

Energy of the photon, ε = hv = hc/λ

= {6.626 x 10-34 Js x 3 x 108 ms-1}/{0.57 x 10-6 m}

= 3.487 x 10-19 J

Therefore, Rate of emission = {25 Js-1}/{3.487 x 10-19 J} = 7.18 x 1019 s-1.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...