Chapter 7, Problem 7.3P

(a)

Interpretation Introduction

Interpretation:

The force exerted by the sail for the spacecraft designed to work by photon pressure which sail was completely absorbing fabric of given area $\text{1}{\text{km}}^{\text{2}}$ and directed a red laser beam with $\lambda 650nm$ has to be calculated.

Concept Introduction:

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\end{array}$

h is Planck’s constant($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$) which relates energy and frequency.

$\text{υ}$ is the speed of particle.

m is the mass of particle.

$\text{λ}$ is the wavelength.

p is the momentum.

The above equation is called de Broglie relation.

The force exerted is identified by multiplying photon momentum with Avogadro’s number.

Explanation of Solution

Given:

Fabric area $\Rightarrow \text{A =}1k{m}^2$

Wavelength $\Rightarrow \lambda \text{ =}650nm$

The force is determined by multiplying the photon momentum with Avogadro’s number. First the momentum is calculated as follows,

$\begin{array}{c}{p}_{photon}=\frac{h}{\lambda }\\ =\frac{6.626\times {10}^{-34}J.s}{650\times {10}^{-9}m}\\ \\ =1.02\times {10}^{-27}kgm{s}^{-1}\\ \\ \left[\begin{array}{c}here,h=Planck\text{'}s\text{constant}\left(6.626\times {10}^{-34}J.s\right)\\ J=kg{m}^2{s}^{-2}\end{array}\right]\end{array}$

$\begin{array}{c}\\ F=\left(N_A{s}^{-1}\right)\times p_{photon}\\ =\left(6.022\times {10}^{23}{s}^{-1}\right)\left(1.02\times {10}^{-27}kgm{s}^{-1}\right)\\ =6.14\times {10}^{-4}N\end{array}$

(b)

Interpretation Introduction

Interpretation:

The exerted pressure by given radiation on the sail for the designed spacecraft has to be calculated.

Concept Introduction:

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\end{array}$

h is Planck’s constant($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$) which relates energy and frequency.

$\text{υ}$ is the speed of particle.

m is the mass of particle.

$\text{λ}$ is the wavelength.

p is the momentum.

The above equation is called de Broglie relation.

The force exerted is identified by multiplying photon momentum with Avogadro’s number.

Force per unit area is called pressure.

Explanation of Solution

Given:

Fabric area $\Rightarrow \text{A =}1k{m}^2$

Wavelength $\Rightarrow \lambda \text{ =}650nm$

The pressure exerted is as follows,

$\begin{array}{c}P=\frac{F}{A}\\ \\ =\frac{6.14\times {10}^{-4}N}{1\times {10}^6{m}^2}\\ \\ =6.14\times {10}^{-10}{\text{N/m}}^{\text{2}}\\ \end{array}$

(c)

Interpretation Introduction

Interpretation:

The time taken by the craft of mass $1kg$ to accelerate speed of $\text{1}{\text{ms}}^{\text{-1}}$ has to be calculated.

Concept Introduction:

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\end{array}$

h is Planck’s constant($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$) which relates energy and frequency.

$\text{υ}$ is the speed of particle.

m is the mass of particle.

$\text{λ}$ is the wavelength.

p is the momentum.

The above equation is called de Broglie relation.

The force exerted is identified by multiplying photon momentum with Avogadro’s number.

Force per unit area is called pressure.

According to Newton’s second law the force is equal to $\text{ma}$.

$\begin{array}{c}\text{F}\text{=}\text{m a}\\ \text{=}\text{m}\frac{v}{t}\\ t=\frac{mv}{F}\\ here,F=force\\ \text{m}\text{=}\text{mass}\\ \text{a}\text{=}\text{acceleration}\\ \text{t}\text{=}\text{time}\text{taken}\\ \text{v}\text{=}\text{velocity}\end{array}$

Explanation of Solution

Given:

Fabric area $\Rightarrow \text{A =}1k{m}^2$

Wavelength $\Rightarrow \lambda \text{ =}650nm$

Speed $\Rightarrow \text{ =}1m{s}^{-1}$

Hence, velocity is equal to $1m{s}^{-1}$

The time taken is as follows,

$\begin{array}{c}\\ t=\frac{mv}{F}\\ =\frac{1kg\times 1m{s}^{-1}}{6.14\times {10}^{-4}N}\\ =1628.7s\\ =1.6287\times {10}^{3}s\\ =0.45h\end{array}$