Physics > Basic Modern Physics > 8.0 Force exerted by a light beam on a surface
Basic Modern Physics
1.0 Photon theory of light
2.0 Characteristics of photon
3.0 Wave Particle Duality
4.0 Emission of electrons
5.0 Photoelectric Effect
5.1 Laws of Photoelectric emission
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
6.0 Radiation Pressure And Force
7.0 Photon Density
8.0 Force exerted by a light beam on a surface
9.0 Early Atomic Structures
10.0 Bohr Model of The Hydrogen Atom
10.1 Radius of Orbit
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
11.0 Energy of electron in the $n^{th}$ orbit
12.0 Basic Definitions
13.0 Atomic Excitation
8.1 Surface exposed to normal radiation
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
Energy incident per unit time $ = IA$
$$\begin{equation} \begin{aligned} {\text{Number of photons emitted per unit time}}\quad N = \frac{{IA}}{{hf}} = \frac{{IA\lambda }}{{hc}} \\ \\\end{aligned} \end{equation} $$
| Case 1. The whole amount of radiation incident is absorbed by the surface. | Case 2. The whole amount of radiation incident is reflected by the surface. | Case 3. Some part of incident radiation is absorbed and rest is reflected by the surface. |
| So, $a = 1$ and $b = 0$. $$\begin{equation} \begin{aligned} {\text{Initial momentum of one photon}} = \frac{h}{\lambda }{\text{ and}} \\ \\\end{aligned} \end{equation} $$ $$\text{Final momentum of one photon} = 0$$ $$\begin{equation} \begin{aligned} {\text{Total change in momentum of one photon}} = \frac{h}{\lambda } \\ \\\end{aligned} \end{equation} $$ Therefore, force on photons = Total change in momentum per unit time $$\begin{equation} \begin{aligned} = N\frac{h}{\lambda } = \frac{{IA\lambda }}{{hc}} \times \frac{h}{\lambda } = \frac{{IA}}{c} \\ \\\end{aligned} \end{equation} $$ Force on plate due to photons (from Newton Thirds Law), $$\begin{equation} \begin{aligned} F = \frac{{IA}}{c} \\ \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} {\text{Pressure}} = \frac{F}{A} = \frac{{IA}}{{Ac}} = \frac{I}{c} \\ \\\end{aligned} \end{equation} $$ | So, $a = 0$ and $b = 1$. $$\begin{equation} \begin{aligned} {\text{Initial momentum of one photon}} = \frac{h}{\lambda }{\text{ and}} \\ \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} {\text{Initial momentum of one photon}} = \frac{h}{\lambda } \\ \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} {\text{Total change in momentum of one photon}} = \frac{{2h}}{\lambda } \\ \\\end{aligned} \end{equation} $$ Therefore, force on photons = Total change in momentum per unit time $$\begin{equation} \begin{aligned} = N\frac{{2h}}{\lambda } = \frac{{IA\lambda }}{{hc}} \times \frac{{2h}}{\lambda } = \frac{{2IA}}{c} \\ \\\end{aligned} \end{equation} $$ Force on plate due to photons(from Newton Thirds Law), $$\begin{equation} \begin{aligned} F = \frac{{2IA}}{c} \\ \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} {\text{Pressure}} = \frac{F}{A} = \frac{{2IA}}{{Ac}} = \frac{{2I}}{c} \\ \\\end{aligned} \end{equation} $$ | So, $0 < a < 1$ and $0 < r< 1$. Force on plate = Force due to absorbed photons (${F_a}$) + Force due to reflected photons (${F_r}$) $$\begin{equation} \begin{aligned} = {F_a} \times (1 - r) + {F_r} \times r \\ = \frac{{IA}}{c}(1 - r) + \frac{{2IAr}}{c} \\ = \frac{{IA}}{c}(1 + r) \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} {\text{Pressure}} = \frac{F}{A} = \frac{I}{c}(1 + r) \\ \\\end{aligned} \end{equation} $$ |
Question 8. (a) How many photons of a radiation of wavelength $\lambda = 5 \times {10^{ - 7}}\ m$ must fall per second on blackened plate in order to produce a force of $$6.62 \times {10^{ - 5}}\ N$$
(b) At what rate will the temperature of plate rise if its mass is $19.86$ $kg$ and specific heat is equal to $2500 J$ ( $kg{K^{ - 1}}$ )?
Solution: (a) If $n$ is the number of photons falling per second on the plate, then the momentum per second of the incident photons is $$\begin{equation} \begin{aligned} P = n \times \frac{h}{\lambda } \\ \\\end{aligned} \end{equation} $$
Since the plate is blackened, all photons are absorbed by it.
$$\begin{equation} \begin{aligned} {\text{Since}} \\ \quad \quad \quad F = \frac{{\Delta P}}{{\Delta t}} = n\frac{h}{\lambda }\quad \Rightarrow \quad n = \frac{{F\lambda }}{h} \\ {\text{or }}\quad {\text{ }}n = \frac{{\left( {6.62 \times {{10}^{ - 5}}} \right)\left( {5 \times {{10}^{ - 7}}} \right)}}{{6.62 \times {{10}^{ - 34}}}} = 5 \times {10^{22}} \\\end{aligned} \end{equation} $$
(b) Energy of each photon = $hc$/$\lambda $
Since $n$ photons fall on the plate per second, the total energy absorbed by the plates in one second is
$$\begin{equation} \begin{aligned} \quad \quad \quad E = n \times \frac{{hc}}{\lambda } = 1986\ J{s^{ - 1}} \\ {\text{i}}{\text{.e, }}\quad {\text{ }}\frac{{dQ}}{{dt}} = 1986\ J{s^{ - 1}} \\ \quad \quad \quad mc\frac{{dT}}{{dt}} = 1986 \\ \; \Rightarrow \quad \quad \frac{{dT}}{{dt}} = \frac{{1986}}{{19.86 \times 2500}} = 4 \times {10^{ - 2}}\ \ \ \ \ {}^0C{s^{ - 1}}\\\end{aligned} \end{equation} $$
