A source produces monochromatic light of frequency \(5.0\times {10}^{14}Hz\) and the power emitted is 3.31 mW. The numbe…
A source produces monochromatic light of frequency \(5.0\times {10}^{14}Hz\) and the power emitted is 3.31 mW. The number of photons emitted per second by the source, on an average is
\({10}^{16}\)
The energy of a single photon is given by the formula \(E=h\nu\) ,where \(h\) is Planck's constant. The value of Planck's constant is
\(h=6.626\times10^{-34}J\cdot s\).
\( E=(6.626\times10^{-34}J\cdot s)(5.0\times10^{14}Hz) \)
\( E=3.313\times10^{-19}J \)
Step 3: Calculate the number of photons emitted per second
The number of photons emitted per second, N, is the total power divided by the energy of a single photon.
\( N=\frac{P}{E} \)
\( N=\frac{3.31\times10^{-3}W}{3.313\times10^{-19}J} \)
\(N\approx9.99\times10^{15}\text{ photons/s} \)
Practice more Board Physics PYQs
See every question on Dual Nature of Radiation and Matter, or browse the full Board question bank.
See all questions on Dual Nature of Radiation and Matter →