Comparison of Isothermal and Adiabatic Processes for an Ideal Gas

Isothermal Process:

1.) In this process temperature remains constant i.e.$(\Delta T= 0)$.

2.) In this process internal energy remains constant i.e. $(\Delta U= 0)$.

3.) This process takes place very slowly.

4.) In this process the system is surrounded by a perfectly conducting material, whose conductivity is infinite.

5.) This process obeys Boyle's law i.e. $(PV= constant)$.

6.) In this process the slope of isothermal curve $=-\frac{P}{V}$

7.) In this process specific heat of gas should be infinite.

Adiabatic Process:

1.) In this process exchange of heat does not take place i.e. $(\Delta Q= 0)$ but temperature changes.

2.) In this process internal energy changes.

3.) This process takes place very rapidly.

4.) In this process the system is surrounded by a perfectly insulating material, whose conductivity is zero.

5.) This process obeys Poisson's law i.e. $(PV^{\gamma} = constant)$.

6.) In this process the slope of adiabatic curve $=- \gamma \frac{P}{V}$

7.) In this process specific heat of gas should be zero.

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Concept of Perfect Gas

Concept of Perfect (ideal) Gas:

An imaginary gas whose properties are similar to the properties of a real gas (a gas whose molecules occupy space and interact with each other) at infinitely low pressure. This imaginary gas is called 'perfect gas' or ideal gas'.

According to the definition, the following properties are imagined in a perfect gas :

(1) It strictly obeys Boyle's law, Charles' law, and the law of pressure under all conditions of temperature and pressure.

(2) Its pressure coefficient and volume coefficient are exactly equal to each other.

(2) Its molecules are infinitesimally small.

(3) There is no force of attraction between its molecules. Obviously, a perfect gas cannot be converted into a liquid or solid state, because a force of attraction is necessary between the molecules in the liquid or the solid state.

In practice, the gases that are difficult to liquefy, such as oxygen, nitrogen, hydrogen, and helium can be considered as perfect, although these are also not ideally perfect.

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Failure of Wave Theory in Explaining Photoelectric Emission Effect

Description of failure of wave theory in explaining photoelectric effect:

Although reflection, refraction, interference, diffraction and polarisation etc. are explained on the basis of wave theory but the laws of photoelectric effect cannot be explained on the basis of the wave theory of light. There are three main reasons for failure:

1.) According to wave theory, as the intensity of incident light increases, incident energy also increases. Therefore, greater is the intensity, greater will be the energy absorbed by the electrons of metal and therefore greater should be the kinetic energy of photoelectrons. From experimental observations, it is clear that the maximum kinetic energy of photoelectrons does not depend on the intensity of incident light.

2.) According to wave theory, photoelectric emission should occur for all the frequencies provided that it has enough energy to emit the electrons from the metal. Although from experimental observation it is clear that if the frequency of incident light is less than the threshold frequency, photoelectrons are not emitted.

3.) The energy carried by the light waves is absorbed by all the electrons and not by a single electron. Therefore, if the intensity of light is less, for the emission of electrons, there should be some time to collect sufficient energy. Although it is clear from experimental observation, the electrons are emitted instantaneously, whatever small be the intensity of light.

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Einstein Photoelectric Emission Effect Law

Description of the Light:

According to the Albert Einstein

The light is consist of small packets or bundles of energy. These packets are called photon. The energy of each photon is $h \nu$.

Theory of Photoelectric Effect:

When the photon of ultraviolet ray is incident on the metal then this energy is used in two parts:

1.) To emit the electron from a metal surface i.e. Work Function. The electron emitted from a metal surface is called the photoelectron.

2.) Remaining energy increases the kinetic energy of the emitted photoelectron. This is the maximum kinetic energy of the photoelectron.

Let us consider

The energy of incident Ultraviolet Photon $E=h\nu$

The work function $W= h \nu_{\circ}$

The maximum kinetic energy of photo electron $K_{max}= \frac{1}{2}mv^{2}_{max}$

According to Einstein's photoelectric effect theory

$E=W+K_{max}$

$h \nu = h \nu_{\circ} + \frac{1}{2}m v^{2}_{max}$

$h \nu - h \nu_{\circ} = \frac{1}{2}m v^{2}_{max}$

$h \left( \nu - \nu_{\circ} \right) = \frac{1}{2}m v^{2}_{max}$

Where
$\nu \rightarrow$ Frequency of incident Ultraviolet ray photon
$\nu_{\circ} \rightarrow$ Threshold frequency to emit the photoelectron from the metal surface
$ v_{max} \rightarrow$ Maximum velocity of emitted photoelectron from the metal surface

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Laws of photoelectric emission

There are the following laws of photoelectric emission:

1.) The rate of emission of photoelectrons from the metal surface is directly proportional to the intensity of the incident light on the metal surface.

2.) The maximum kinetic energy of photoelectrons does not depend on the intensity of incident light.

3.) If the frequency of incident light is less than the threshold frequency then no photoelectrons will come out of the surface.

4.) If the frequency of incident light is equal to the threshold frequency then electrons will come out of the surface but the kinetic energy of emitted photoelectrons will be zero.

5.) As the frequency of incident light increases greater than the threshold frequency, the maximum kinetic energy $K_{max}$ of photoelectrons emitted from the metal surface also increases.

6.) The value of threshold frequency depends upon the nature of the metal surface and its value is different for different metals.

7.) There is no time lag in the emission of photoelectrons to the metal surface i.e., as the light of proper from frequency is made to the incident on the metal surface, photoelectrons immediately come out of the surface within no time.

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Expression for fringe width in Young's double slit experiment

Expression for fringe's width:

Let us consider two wave from slit $S_{1}$ and $S_{2}$ superimpose on each other and form interfernece patteren on the screen. The distance between the two slits is $d$ and distance between slit to screen is $D$. Now take a $n^{th}$ fringe from the centre $O$ of the screen which is at distance $y_{n}$.

So the path difference between the rays

$\Delta x = S_{2}P- S_{1}P \quad(1)$

In $\Delta S_{1}PM$

$S_{1}P^{2}=S_{1}M^{2}+PM^{2} \quad(2)$

From figure:

$S_{1}M =D$
$PM= y_{n}- \left(\frac{d}{2}\right) $

Now subtitute these values in equation $(2)$, then

$S_{1}P^{2}=D^{2}+ \left( y_{n}- \frac{d}{2} \right)^{2} \quad(3)$

In $\Delta S_{2}PN$

$S_{2}P^{2}=S_{2}N^{2}+PN^{2} \quad(4)$

From figure:

$S_{2}N =D$
$PN= y_{n} + \left(\frac{d}{2}\right) $

Now subtitute these values in equation $(4)$, then

$S_{2}P^{2}=D^{2}+ \left( y_{n} + \frac{d}{2} \right)^{2} \quad(5)$

Now subtract the equation $(3)$ in equation $(5)$

$S_{2}P^{2} - S_{1}P^{2} = \left[D^{2}+ \left( y_{n} + \frac{d}{2} \right)^{2}\right] - \left[ D^{2}+ \left( y_{n}- \frac{d}{2} \right)^{2} \right] $

$S_{2}P^{2} - S_{1}P^{2} = D^{2}+ \left( y_{n} + \frac{d}{2} \right)^{2} - D^{2} - \left( y_{n}- \frac{d}{2} \right)^{2} $

$S_{2}P^{2} - S_{1}P^{2} = \left( y_{n} + \frac{d}{2} \right)^{2} - \left( y_{n}- \frac{d}{2} \right)^{2} $

$S_{2}P^{2} - S_{1}P^{2} = y^{2}_{n} + \left(\frac{d}{2}\right)^{2} +2 y_{n} \left(\frac{d}{2}\right) - y^{2}_{n} - \left(\frac{d}{2}\right)^{2} +2 y_{n} \left(\frac{d}{2}\right) $

$S_{2}P^{2} - S_{1}P^{2} = y^{2}_{n} + \left(\frac{d}{2}\right)^{2} + y_{n} d - y^{2}_{n} - \left(\frac{d}{2}\right)^{2} + y_{n} d $

$S_{2}P^{2} - S_{1}P^{2} = y_{n} d + y_{n} d $

$ \left( S_{2}P + S_{1}P \right) \left( S_{2}P - S_{1}P \right) = 2y_{n} d$

$ \left( S_{2}P + S_{1}P \right) \Delta x = 2y_{n} d \quad \left\{from \: equation\: (1)\right\} \qquad(6)$

In this experiment, the distance between slit and screen i.e. $D$ is much greater than the distance between the slit i.e. $d$ so angle $\theta$ will be very small. So from figure

$S_{1}P = S_{2}P = D$

Substitute this value in the equation $(6)$, therefore equation $(6)$ can be written as

$ 2D \Delta x = 2y_{n} d $

$D \Delta x = y_{n}$

$\Delta x = \frac{y_{n} d}{D} \quad (7) $

This is equation of path differnce between two wave from slit $S_{1}$ and slit $S_{2}$.

Condition For Bright Fringe:

The path difference for bright fringe is

$\Delta x= n \lambda \qquad(8)$

From equation $(7)$ and equation $(8)$

$n \lambda = \frac{ d}{D} $

$y_{n} = \frac{n \lambda D}{d} \quad(9)$

This is the equation for distance of $n^{th}$ bright fringe. Now the distance of $(n+1)^{th}$ bright fringe.

$y_{n+1} = \frac{\left( n+1 \right) \lambda D}{d} \quad(10)$

The dark fringe lie between the two consecutive bright fringe. So width of the dark fringe

$\beta = y_{n+1} - y_{n}$

Now subtitute the value of $y_{n+1}$ and $y_{n}$ in the above equation $(9)$ and equation $(10)$, then

$\beta = \frac{\left( n+1 \right) \lambda D}{d} - \frac{n \lambda D}{d}$

$\beta = \frac{\lambda D}{d} \quad(11)$

This is the equation of width of dark fringe.

Condition For Dark Fringe:

The path difference for dark fringe is

$\Delta x = (2n-1) \lambda \quad(12)$

From equation $(7)$ and equation $(12)$

$ (2n-1) \lambda = \frac{y_{n} d}{D} $

$y_{n} = \frac{(2n-1) \lambda D}{d} \quad(13)$

This is the equation for distance of $n^{th}$ dark fringe. Now the distance of $(n+1)^{th}$ dark fringe.

$y_{n+1} = \frac{\left[ 2 (n+1)-1) \right] \lambda D}{d} $

$y_{n+1} = \frac{(2n+1)\lambda D}{d} \quad(14)$

The bright fringe lie between the two consecutive dark fringe. So width of the bright fringe

$\beta = y_{n+1} - y_{n}$

Now subtitute the value of $y_{n+1}$ and $y_{n}$ in the above equation $(13)$ and equation $(14)$, then

$\beta = \frac{(2n+1)\lambda D}{d} - \frac{(2n-1) \lambda D}{d}$

$\beta = \frac{\lambda D}{d} \quad(15)$

This is the equation of width of bright fringe.

The equation $(11)$ and equation $(15)$ shows that the width of bright fring and dark fringe is same.

Alternative method to find the path differnce of two wave in Young's double-slit Experiment

Let us consider: Two wave from slit $S_{1}$ and $S_{2}$ superimpose on each other and form interfernece patteren on the screen.

The distance between the two slits is $\rightarrow d$

The distance between slit to screen is $\rightarrow D$

The distance of $n^{th}$ fringe from the centre $O$ of the screen is $\rightarrow y_{n}$.

So the path difference between the two rays is

$\Delta x = S_{2}M \quad(1)$

Now from figure, In $\Delta S_{1}S_{2}M$

$sin \theta = \frac{S_{2}M}{S_{1}S_{2}} \quad(2)$

In $\Delta PNO$

$tan \theta = \frac{PO}{NO} \quad(3)$

The distance between slit and screen i.e. $D$ is much greater than the distance between the slit i.e. $d$ so angle $\theta$ will be very small i.e. $sin \theta \approx tan \theta \approx \theta $. Now equate the equation $(2)$ and equation $(3)$

$\frac{PO}{NO} = \frac{S_{2}M}{S_{1}S_{2}} \quad(4)$

Now the value from the above figure

$\frac{y_{n}}{D} = \frac{\Delta x}{d} $

$\Delta x = \frac{y_{n} d}{D} $

This is the equation of the path difference between two wave from slit $S_{1}$ and $S_{2}$.

Now the derivation for the condition for Bright and Dark fringe can be used here as it is given above.

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