Chapter 26, Problem 22E

(a)

To determine

The number of photons emitted per second.

Answer to Problem 22E

Thenumber of photons emitted per secondis $\underset{\_}{2.76\times {10}^{18}\text{/s}}$.

Explanation of Solution

Given that the power is $1\text{W}$ and average wavelength is $550\text{nm}$.

Write the expression for energy having $n$ photons of wavelength.

  $E=\frac{nhc}{\lambda }$        (I)

Here, $n$ is the number of photons, $E$ is the energy, $h$ is the Planck’s constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the photon.

Write the expression for energy in terms of power.

  $E=pt$        (II)

Here, $P$ is the power and $t$ is the time.

Use Equation (II) in (I) and solve for $n$.

  $\begin{array}{c}pt=\frac{nhc}{\lambda }\\ n=\frac{pt\lambda }{hc}\end{array}$        (III)

Write the expression for number of photons emitted per second.

  $N=\frac{n}{t}$        (IV)

Here, $N$ number of photons emitted per second.

Solve Equation (IV) for $n$.

  $n=Nt$        (V)

Use Equation (V) in (III).

  $\begin{array}{c}Nt=\frac{pt\lambda }{hc}\\ N=\frac{p\lambda }{hc}\end{array}$        (VI)

Conclusion:

Substitute $6.62\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $550\text{nm}$ for $\lambda$, and $1\text{W}$ for $P$ in Equation (VI) to find the number of photons emitted per second.

  $\begin{array}{c}N=\frac{\left(1\text{W}\right)\left(550\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{\left(6.62\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =2.76\times {10}^{18}\text{/s}\end{array}$

Therefore, the number of photons emitted per secondis $\underset{\_}{2.76\times {10}^{18}\text{/s}}$.

(b)

To determine

The distance from the eye to the candle.

Answer to Problem 22E

The distance from the eye to the candleis $\underset{\_}{892\text{m}}$.

Explanation of Solution

Given that the threshold intensity is ${10}^{-7}\text{W}\cdot {\text{m}}^{-2}$.

Write the expression for intensity in terms of power.

  $I=\frac{P}{A}$        (VII)

Here, $I$ is the threshold intensity, $A$ is the area, and $P$ is the power.

The expression for area of a sphere is as follows:

  $A=4\pi r^2$        (VIII)

Here, radius is $r$.

Use Equation (VIII) in (VII) and solve for $r$.

  $\begin{array}{c}I=\frac{P}{4\pi r^2}\\ r^2=\frac{P}{4\pi I}\\ r=\sqrt{\frac{P}{4\pi I}}\end{array}$        (IX)

Conclusion:

Substitute ${10}^{-7}\text{W}\cdot {\text{m}}^{-2}$ for $I$, and $1\text{W}$ for $P$ in Equation (IX) to find the distance from the eye to the candle.

  $\begin{array}{c}r=\sqrt{\frac{1\text{W}}{4\pi \left({10}^{-7}\text{W}\cdot {\text{m}}^{-2}\right)}}\\ =892\text{m}\end{array}$

Therefore, the distance from the eye to the candleis $\underset{\_}{892\text{m}}$.

(c)

To determine

The number of photons per square meter per second.

Answer to Problem 22E

The number of photons per square meter per secondis $\underset{\_}{2.76\times {10}^{11}{\text{/m}}^2\text{s}}$.

Explanation of Solution

From Equation (VI) in part (a) the expression for number photons emitted per second is expressed as:

  $N=\frac{p\lambda }{hc}$

The expression for the number of emitted per square meter per second is,

  $N=\frac{p\lambda }{Ahc}$        (X)

From Equation (VIII) in part (b) the expression for area is,

  $A=4\pi r^2$

Use Equation (VIII) in (X).

  $N=\frac{p\lambda }{\left(4\pi r^2\right)hc}$        (XI)

Conclusion:

From part (b) the distance from the eye to the candle is $892\text{m}$.

Substitute $6.62\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $550\text{nm}$ for $\lambda$, $892\text{m}$ for $r$, and $1\text{W}$ for $P$ in Equation (XI) to find the number of photons per square meter per second.

  $\begin{array}{c}N=\frac{\left(1\text{W}\right)\left(550\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{\left(4\pi {\left(892\text{m}\right)}^2\right)\left(6.62\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =2.76\times {10}^{11}{\text{/m}}^2\text{s}\end{array}$

Therefore, the number of photons per square meter per secondis $\underset{\_}{2.76\times {10}^{11}{\text{/m}}^2\text{s}}$.