Question

A 250 W electric bulb of 80% efficiency emits a light of 6626 \AA wavelength. The number of photons emitted per second by the lamp is ($h = 6.626 \times 10^{-34}$ Js)

• A. $1.42 \times 10^{17}$
• B. $2.18 \times 10^{16}$
• C. $6.66 \times 10^{20}$
• D. $2.83 \times 10^{16}$
• E. $4.25 \times 10^{16}$

Hint:

When the wavelength is given in Angstroms ($X$ \AA), and Planck's constant is given as $6.626$, look for numerical cancellations! Here, $6626$ and $6.626$ simplify the math significantly.

Answer 1

The Correct Option is C

Concept: The power of a light source is the total energy emitted as light per unit time. According to Planck's quantum theory, light consists of discrete packets of energy called photons.
• Energy of one photon ($E$): $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
• Useful Power ($P_{out}$): $P_{out} = \text{Total Power} \times \text{Efficiency}$.
• Photon Emission Rate ($n$): $P_{out} = n \times E \implies n = \frac{P_{out}}{E}$.

Step 1: Calculate the energy of a single photon. Given $\lambda = 6626 \text{ \AA} = 6626 \times 10^{-10} \text{ m}$. \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6626 \times 10^{-10}} \] \[ E = \frac{6.626 \times 3 \times 10^{-26}}{6.626 \times 10^{-7}} = 3 \times 10^{-19} \text{ J} \]

Step 2: Calculate the number of photons emitted per second. Useful power $P_{out} = 250 \text{ W} \times 0.80 = 200 \text{ J/s}$. \[ n = \frac{P_{out}}{E} = \frac{200}{3 \times 10^{-19}} \] \[ n = 66.66 \times 10^{19} = 6.66 \times 10^{20} \text{ photons/s} \]