Chapter 6, Problem 128AP

Interpretation Introduction

Interpretation:

The number of photons emitted by a light bulb per second is to be calculated for the given wavelength of $550\text{ nm}\text{.}$

Concept introduction:

The energy of a photon can be expressed as

$E=\text{h}\nu \text{or}E=\frac{\text{hc}}{\lambda }$

Here, $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, E is the energy of a photon, $\lambda$ represents the wavelength, and $\nu$ represents the frequency.

The relationship between meters and nanometres can be expressed as $1{\text{ m =10}}^9\text{ nm}$.

To convert nanometers to meters, the conversion factor is $\frac{1\text{ nm}}{1\times {10}^{-9}\text{ m}}$.

The relationship between $\text{W}$ and $\text{J/s}$ can be expressed as $1\text{ W}=1\text{ J/s}$.

To convert $\text{W}$

to $\text{J/s}$, the conversion factor is $\frac{1\text{ J/s}}{1\text{ W}}$.

The number of photons emitted per second can be evaluated as

$\text{Number of photons}=\frac{\text{Eenergy per second}}{\text{Energy of light}}$