Question

A blue lamp emits light of mean wavelength 4500\({Å}\). The lamp is rated at 150 W and 8% efficiency. Then the number of photons are emitted by the lamp per second.

• A. \( 27.17 \times 10^{18} \)
• B. \( 17.17 \times 10^{18} \)
• C. \( 27.17 \times 10^{15} \)
• D. \( 54 \times 10^{16} \)

Hint:

When calculating the number of photons emitted by a lamp, use the relation between power, photon energy, and the efficiency of the system. Photon energy can be calculated using Planck's constant and the wavelength.

Answer 1

The Correct Option is A

To find the number of photons emitted by the lamp per second, we start with the given information:

• Wavelength of the light, \( \lambda = 4500 \text{ Å} = 4500 \times 10^{-10} \text{ m} \)
• Power of the lamp, \( P = 150 \text{ W} \)
• Efficiency of the lamp, \( \eta = 8\% = 0.08 \)

Firstly, calculate the energy of a single photon using the equation:

\( E = \frac{hc}{\lambda} \)

where:

• \( h = 6.626 \times 10^{-34} \text{ J s} \) (Planck's constant)
• \( c = 3 \times 10^8 \text{ m/s} \) (speed of light)

Substituting the values:

\( E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4500 \times 10^{-10}} \)

Calculating \( E \):

\( E = \frac{19.878 \times 10^{-26}}{4500 \times 10^{-10}} \approx 4.4173 \times 10^{-19} \text{ J} \)

The useful power output of the lamp is given by:

\( P_{\text{useful}} = \eta \times P = 0.08 \times 150 \text{ W} = 12 \text{ W} \)

Next, calculate the number of photons emitted per second, \( n \), using the equation \( P_{\text{useful}} = nE \):

\( n = \frac{P_{\text{useful}}}{E} = \frac{12}{4.4173 \times 10^{-19}} \)

Evaluating \( n \):

\( n \approx 27.17 \times 10^{18} \) photons per second

Thus, the number of photons emitted by the lamp per second is \( 27.17 \times 10^{18} \).

Answer 2

We are given that the blue lamp emits light with a wavelength of \( \lambda = 4500 \, {Å} = 4500 \times 10^{-10} \, {m} \), and the lamp is rated at \( 150 \, {W} \) with 8% efficiency. 
Step 1: The power output in terms of the energy of photons is given by the equation:

\( P = \frac{E_{{photon}} \cdot N}{t} \)

where \( P \) is the power output, \( E_{{photon}} \) is the energy of each photon, \( N \) is the number of photons, and \( t \) is the time. The energy of a photon is given by:

\( E_{{photon}} = \frac{h c}{\lambda} \)

where \( h = 6.626 \times 10^{-34} \, {J} \cdot {s} \) (Planck's constant), \( c = 3 \times 10^8 \, {m/s} \) (speed of light), and \( \lambda = 4500 \times 10^{-10} \, {m} \). Substituting the known values:

\( E_{{photon}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4500 \times 10^{-10}} = 4.42 \times 10^{-19} \, {J} \)

Step 2: The total energy output of the lamp per second (considering 8% efficiency) is:

\( {Energy per second} = 0.08 \times 150 = 12 \, {J/s} \)

Step 3: Now, the number of photons emitted per second is:

\( N = \frac{{Energy per second}}{E_{{photon}}} = \frac{12}{4.42 \times 10^{-19}} = 2.72 \times 10^{19} \)

Thus, the number of photons emitted by the lamp per second is \( 27.17 \times 10^{18} \).