Showing posts with label Dual Nature. Show all posts
Showing posts with label Dual Nature. Show all posts

Problems and solutions on de Broglie's Hypothesis

Problem and solution

If 10,000 voltage is applied across an x-ray tube, what will be the ratio of the de  Broglie wavelengths of incident electron to that of the x-ray produced?

To solve this problem, we have to write the wavelength of the electron in terms of the de Broglie concept. Here in the place of the momentum we can write applied potential and the corresponding energy. Similarly to right the wavelength of the x-ray, we can use plank’s quantum concept as shown below.




Problems and solutions

Here we are going to solve two problems. In the first problem and alpha particle and the proton are passed through same magnetic field which is perpendicular to the velocity vectors. If both the charger particles are having the same radius, what is the ratio of their de Broglie wavelengths?

We can solve this problem by first of all understanding that the force due to the magnetic field on the charged particle provides the necessary centripetal force so that they takes a circular path. From that it can be concluded that the momentum of the particle is directly proportional to charge of the particle as well as the radius of the particle. Anyway in this problem being the radius is same, we can say momentum means directly proportional to the charge of the particle itself.

As per this concept, wavelength is inversely proportional to momentum that means wavelength is inversely proportional to the charge of the given particles.



The second problem is also solved basing on the same concept as shown above.

Problem and solution

Photons of energy 4.25 electron volt and 4.7 electron volt are allowed to incident on to metal surface S. If the maximum kinetic energies between them are having a difference of 1.5 electron volts and the they are wavelengths are in the ratio of 1:2, find the work functions of the two different metals?

We can express de Broglie wavelength in terms of the kinetic energy. Basing on this we can say that the ratio of the day Broglie wavelength is inversely proportional to Squire root of ratio of their respective kinetic energies. Taking this concept into consideration we can solve the problem as shown below.




Problem and solution

de Broglie wavelength of the proton accelerated  through a potential difference of 100 V is given. If a alpha particle is accelerated through same potential difference, what will be its wavelength?

According to de Broglie hypothesis, we know that the wavelength of a particle is the ratio of Planck’s constant to the momentum of the particle. We can express the momentum in terms of kinetic energy. Further we can express the kinetic energy in terms of the applied potential. By substituting the given data in that equation, we can solve the problem as shown below.




Problem and solution

A light particle of a certain mass at rest explodes into two particles having masses in the ratio of 2:3 . What is the corresponding ratio of the de Broglie wavelengths?

As the particle is in the state of rest, its initial momentum is equal to 0. The explosion is happened due to internal forces and hence law of conservation of momentum is very much valid here. According to this law the initial momentum of the system is equal to the final momentum when no external forces are acting on the system. As the momentum is as well as the Planck’s constant are same, the wavelengths are also going to be the same.



Problem and solution

What will be the wavelength of electron having energy of hundred electron volts?

We can solve this problem basing on dual nature of the particle. Here the particles shall be having a wave nature and hence it shall have certain wavelength which is equal to the ratio of Planck’s constant to momentum of the particle. We can express the momentum in terms of kinetic energy and the kinetic energy can be further expressed in terms of the applied voltage.

By substituting the data in the given problem we can solve the problem as shown below.



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De Broglie Hypothesis and Wave Nature of Particle

Particle nature of light

Photoelectric effect of light can be explained only when we assume that light these interacting with the matter like a particle. These wave packets are called quanta and they have fixed amount of energy. The energy particles are also called photons and they do not have the rest mass. It can be confirmed that when the light is interacting with the matter, it behaves like a particle.

With the increase of intensity, there will be more number of photons are available in the given light. These photons are electrically neutral and they are not deflected in electric and magnetic fields. When photon collides with a particle, its total energy and momentum are always conservative.

De Broglie Hypothesis

To explain the properties of the light like interference, diffraction and polarization, we shall depend on wave nature of the light. Simultaneously to explain the properties like photoelectric effect and Compton effect, light shall have particle nature.

As the same light is exhibiting both the set of properties, it shall also have dual nature.
In the universe the energy is broadly in the format of matter and radiation. There is a basic concept in the nature which tells that nature loves symmetry. We see so many things in the nature which are symmetrical.

We can explain the concept of symmetry easily. We know that the solar system is a macroscopic system where the sun is at the Center and the planets revolving around it. We also know that the item is a microscopic system where the nucleus is at the Center and the electrons are revolving around it. So we can see a broader symmetry in the design of the nature.
It is already proved that light is having dual nature. It travels like a wave and it interacts like a particle. We know that in the nature both light and the particles are forms of energy’s. As one form of energy is having dual nature and as nature loves symmetry, it is understood that the matter is also having dual nature.

It means to say that all the particles that we see are also having the wave nature. It means the electrons, the protons are not only having the behavior of the particles, and they can also behave like the waves under suitable conditions. This concept is called dual nature of matter. When the particles are travelling like waves they are called as matter waves.

As per the day Broglie’s concept the wavelength of the particle is equal to the ratio of Planck’s constant to the momentum of the particle. Experimentally it is proved that the particle can have wave nature.



So it can be concluded that electrons can behave like waves and can undergo diffraction. We can express the momentum in terms of kinetic energy as well as the applied potentiality as shown below.



We can also express the wavelength of the particle in terms of absolute temperature. It is based down the total energy particle can have for a particular temperature. This equation is derived basing on kinetic theory of gases.




In the above diagram we are also having a problem where the applied potential is given for an electron and we are supposed to calculate the corresponding wavelength of the electron. By converting the momentum into the kinetic energy and further in terms of the voltage with can solve the problem as shown above.


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Dual Nature of Radiation and Matter Complete Lesson

Dual nature of radiation and matter is a concept that can be explained basing on modern physics. In this chapter first photo electric effect is explained. There is emission of photo electrons from the metal surface when a light of suitable frequency incidents on it. It is called photo electric effect. 

In this chapter it is also explained that the particles like electrons and protons can also exhibit wave nature. This is called dual nature of the particle. In the following the links we have given all the relevant information of dual nature of radiation and matter.









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Problems and Solutions on Einstein's Photo Electric Equation

The phenomena of emission of electrons from a metal surface when light of suitable frequency incident on a metal surface is called photo electric effect.The incident light is a stream of photons and each photon has a fixed amount of energy.The incident energy is used to remove the electron from the metal surface and further will acquire some kinetic energy.This energy distribution is explained by Einstein  and the corresponding equation is called Einstein's photo electric equation.Here we are going to solve some problems basing on this concept.

Problem and solution

If a light of frequency 1.5 times threshold frequency incidents on metal surface, electrons are ejected with a certain velocity. If the incident frequency is halved and the intensity is doubled, what will be the velocity with which electrons are going to be ejected ?

The incident frequency shall be at least equal to threshold frequency for the emission of photo electrons. When the incident frequency is reduced to half which is just equal to threshold frequency, it becomes less than threshold frequency. And hence there is not going to be any photo electric effect. Increase of the intensity of light is not going to help here. It is simply because if you increase the intensity, more number of photons were ejected but each photon will have same frequency that is less than threshold frequency. Hence there is no photo electric effect in this case.




Problem and solution

The stopping potential of a metal surface is given as 9 V . What is the maximum speed of the rejected electron in this case?

We can solve this problem basing on the simple concept. When the electron is rejected from the cathode it will how kinetic energy and it will be moving towards the anode. When we apply reverse potentiality, at a particular potential photo electric current stops and that potentially is called stopping potential. So at the stopping potential, electron is having a energy which is equal to kinetic energy of the electron and opposite indirection. And hence both of them are able to be cancelled each other and their electron is getting stopped.

Taking this into consideration we can equate both the energies and solve the problem as shown in the first part of the given diagram.

Problem and solution

The page attached below is having another problem.

A graph is drawn taking the potential on x-axis and the photo electric current on y-axis. Three different situations were given and you are asked to identify, which of the following your options are correct?

You can analyse a question and comment at the end of the post for any kind of clarifications.


Problem and solution

A graph is drawn taking the applied potential on x-axis and the corresponding photo electric current on y-axis. Find the relation between the relation between the wavelengths of the incident light?

From the graph it is very clear that stopping potential for the second wavelength is more than that of the stopping potential of the first wavelength. We know that stopping potentially is directly proportional to incident frequency, when the incident frequency is more than the threshold frequency. Therefore as the second stopping potentially is more we can say the second incident frequency is more. As the frequency is reciprocal to wave length, second wavelength is less than that of the first wavelength.



Problem and solution

A graph is drawn taking the incident frequency on x-axis and the corresponding stopping potential on y-axis. What is the threshold wavelength of the light used ?

From the graph it is very clear that photo electric current starts from a certain frequency and that certain frequency is called the threshold frequency. Using the relation between velocity and the frequency of the light we can calculate the wavelength as shown below.



The above attached paper is having one more small problem.

Problem and Solution

If the velocity of the most energetic electron emitted is doubled when the incident frequency is doubled, what is the work function of the metal surface?

We can solve this problem basing on photo electric effect equation. According to the equation the incident energy is equal to sum of work function and the corresponding kinetic energy of the emitted electron. By applying the given data in that equation we can solve the problem as shown above.


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Photo Electric Effect Problems and Solutions

The phenomenon of emission of electrons from a metal surface when a light of suitable frequency incident of the metal surface is called photo electric effect.

Problem and solution

A metal of work function of 4 electron volts is exposed to radiation of the wavelength 140 nm. What is the corresponding stopping potential?

We can solve this problem basing on the very definition of energy. It is written basing on plank’s quantum concept. We can further write photo electric equation to solve the problem as shown below.




Problem and solution

On the cathode of a photoelectric cell two different wavelengths of light are allowed to incident. The work function of the metal surface is given. A uniform magnetic field of known magnetic field strength is applied perpendicular to the cell. Find the radius of circular path described by the photo electrons?

To solve this problem, first of all we have to analyse whether the incident light is sufficient enough energy for the emission of photo electrons or not.

It is mathematically observed that photo electric effect is possible only with the first wavelength but not with the second wavelength. It is simply because the wavelength and the second case is more than the threshold wavelength.

We shall also know that whenever a magnetic field is applied perpendicular plane, the charged particle like electron takes a circular path and the necessary centripetal forces provided by the force due to the magnetic field on the charger particles.

The problem is solved as shown below.



Problem and solution

If the wavelength of the incident radiation changes from one value to other value, the corresponding kinetic energy emitted by the photo electrons also changes from one value to other value. What is the work function of the metal surface?

In solving this problem, we can use photo electric equation. According to this equation the total energy supplied by the photo is equal to sum of work function and the kinetic energy of the electron. By applying this condition for both the cases and by simplifying the equation we can derive the equation further work function as shown below.



Problem and solution

For a certain metal threshold frequency is given. When the incident frequency is doubled the threshold frequency, electrons comes out with a certain velocity. If the incident frequency is five times threshold frequency, what is the velocity with which the electrons come out?

We can solve this problem also basing on photo electric equation. We know that the kinetic energy of the electron emitted is equal to the difference between the energy of the incident photon and the work function. By substituting this condition for both the cases, cases that are given in the problem, we can solve it.




Problem and solution

Photo electric effect from a metallic surface is observed from two different frequencies where first frequency is greater than that of the second frequency. If the ratio of the maximum kinetic energy emitted by the photo electrons in both the cases is given, find the threshold frequency of the metal surface?

We know that the incident energy of the photon is equal to sum of work function and the kinetic energy. Applying this concept for both the frequencies that are given, we can calculate the threshold frequency as shown below.



Problem and solution

A photon of known energy ejects photo electrons from metal surface with a certain work function. If this emitted electron enters into a uniform magnetic field of known induction perpendicular to the field, what is the radius of the circular path taken by it?

As the total energy of the photon is equal to sum of work function and kinetic energy, we can write the equation for the velocity of the ejected electron as shown below. This electron that is entering into the magnetic field perpendicularly shall experience a centripetal force and shall take a circular path. The necessary centripetal forces provided by the force experienced by the electron due to the magnetic field.



Problem and solution

When a light of known wavelength incident on a photoelectric surface, the velocity of the fastest electron emitted is  V . If the incident wavelength is 75% of the initial wavelength, what will be the velocity of the emitted electron?

We can solve this problem basing on photo electric equation. According to this equation the total energy supplied by the light in the form of photon, is equal to sum of work function as well as the kinetic energy of emitted electron. By substituting the given values in this equation, we can get to equations. By solving that to equations we can identify the relation between the last of the electron in the second case with respect to the velocity of the electron in the first case.



Problem and solution

Here we are going to solve a simple problems basing on photo electric effect.

When a radiation of certain wavelength is incident on a metallic surface, the stopping potentially is found to be 4.8 V. If the same surface is illuminated by the radiation of double the wavelength, the stopping potential is found to be 1.6 V. What is the threshold wavelength of the surface?

We can solve this problem basing on Einstein’s  photo electric equation. According to this equation the total energy supplied by the photon is equal to sum of work function and kinetic energy. By applying the given data in this equation in two different cases and by simplifying the equation is we can solve the problem as shown below.



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Einstein’s Photo Electric Equation and Applications

Einstein’s photo electric equation explains photo electric effect and its further properties successfully. Wave theory of light has failed to explain concept of photo electric effect. Hence plank’s quantum concept is taken into consideration to explain this property. According to this property light is emitted by a source not continuously but discreetly in the form of wave packets called quanta.

When the light strikes a surface, number of photons are striking the surface. More the intensity, more the number of photons. Each photon will have certain energy, basing on the frequency of incident light. Each photon will pass all its energy to a electron on the surface of the metal. They electron is associated with the metal and it needs some energy to release itself from the valency orbit.

When the energy given by the photon is sufficient to release the electron from the metal surface, the electrons are emitted from the metal surface. The number of the electrons emitted from the metal surface depends on the number of the photons that are striking the metal surface. If the intensity of the light is more, there is availability of more number of photons and hence more photo electric current is emitted. That is the reason why intensity and the photo electric current are proportional to each other.

The electrons in the valency orbit need a minimum energy to get released itself from the metal surface. The incident frequency shall be able to give at least this much of energy to release the electron. This minimum frequency is called threshold frequency. This minimum required energy for the release of the electron is called work function. The energy given by the photon is first used to release the electron from the metal surface. If there is any further energy that is given by the photon, the electron uses that energy and moves with the kinetic energy.

According to photo electric equation, the entire energy supplied by the photon is used first to release the electron from the metal surface and then to move the electron from the cathode towards the anode. First the energy has to be supplied to release the electron from the metal surface and only when there is an extra energy it can be used to move the electron. Supplying of energy equal to work function is the precondition for the emission of photo electrons.

We can derive the equation for the maximum kinetic energy of by a electron as shown below.



If intensity of light is more, more photons are available and each photon will go and give its energy to the electrons. And hence there is more electrons emitted and hence there will be more photo electric current. The number of the electrons emitted is independent of frequency. But whether the current is able to be emitted are not is a dependent of frequency. The incident frequency and its corresponding energy shall be sufficient to release the electron from the metal surface. That is the energy of the frequency shall be more than the work function for the photo electric effect to happen.

Application

A monochromatic radiation incident on a metal surface of work function W, maximum kinetic energy of the emitted electrons is K. If the frequency of the incident radiation is n times the initial frequency and then how its kinetic energy will be affected?

We can write photo electric equation to solve this problem. We know that the total energy of the photon is equal to the sum of work function and the kinetic energy of the electron. Basing on that, we can write the equation for the kinetic energy in both the cases with their respective frequencies as shown below. Simply buy solving this to equations we can get the answer.




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