Question

A 60 W source emits monochromatic light of wavelength 662.5 nm. The number of photons emitted per second is

• A. 2 × 10²⁰
• B. 5 × 10²⁶
• C. 2 × 10²⁹
• D. 5 × 10¹⁷

Answer 1

The Correct Option is A

Given Information: 
Power of source, \(P = 60\,W\)
Wavelength of monochromatic light, \(\lambda = 662.5\,nm = 662.5 \times 10^{-9}\,m\)

Step-by-Step Explanation:

Step 1: Energy per photon clearly calculated:

The energy of each photon (\(E\)) is given by:

\[ E = \frac{hc}{\lambda} \]

where \(h = 6.626\times10^{-34}\,Js\), and \(c = 3\times10^8\,m/s\).

Substitute clearly:

\[ E = \frac{6.626\times10^{-34}\times 3\times10^8}{662.5\times10^{-9}} \]

Simplifying clearly,

\[ E \approx \frac{19.878\times10^{-26}}{662.5\times10^{-9}} \approx 3\times10^{-19}\,J \]

Step 2: Calculate clearly the number of photons emitted per second:

The number of photons emitted per second (\(n\)) is given by power divided by energy per photon:

\[ n = \frac{P}{E} = \frac{60}{3\times10^{-19}} = 2\times10^{20}\,s^{-1} \]

Final Conclusion:
The number of photons emitted per second is clearly \(2\times10^{20}\,s^{-1}\).

Answer 2

The power of the source is given by \( P = 60 \, \text{W} = 60 \, \text{J/s} \). The energy of one photon is given by the equation: \[ E = \frac{hc}{\lambda} \] where: 
\( h \) is Planck's constant \(= 6.626 \times 10^{-34}\ \text{J.s},\)
\( c \) is the speed of light \( = 3.0 \times 10^8 \, \text{m/s} \),
\( \lambda \) is the wavelength of the light \( = 662.5 \, \text{nm} = 662.5 \times 10^{-9} \, \text{m} \).

Substituting the values: \[ E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{662.5 \times 10^{-9}} = 3.0 \times 10^{-19} \, \text{J} \] Now, the number of photons emitted per second is: \[ \text{Number of photons} = \frac{P}{E} = \frac{60}{3.0 \times 10^{-19}} = 2 \times 10^{20} \] Thus, the number of photons emitted per second is \( 2 \times 10^{20} \).