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Dual Nature of Radiation and Matter | Physics - Short Answer Questions | 12th Physics : UNIT 8 : Dual Nature of Radiation and Matter

Chapter: 12th Physics : UNIT 8 : Dual Nature of Radiation and Matter

Short Answer Questions

Physics : Dual Nature of Radiation and Matter: Book Back Important Questions, Answers, Solutions: Short Answer Questions

II Short Answer Questions

 

1. Why do metals have a large number of free electrons?

In metals, outer most electrons are loosly bound with nucleus.

Even at room temperature, these electrons are freely moving with in the metal. These electrons are called free electrons. Hence metals having large number of free electrons.


2. Define work function of a metal. Give its unit.

The minimum energy needed for an electron to escape from the metal surface is called work function of that metal (ϕo).

Its unit is Joule (or) eV.


3. What is photoelectric effect?

The ejection of electrons from a metal plate when illuminated by light or any other electromagnetic radiation of suitable wavelength (or) frequency is called photoelectric effect.


4. How does photocurrent vary with the intensity of the incident light?

The photocurrent increases as intensity of light increase. Since number of electrons emitted per second from the surface is directly proportional to the intensity of the light.


5. Give the definition of intensity of light and its unit.

According to quantum concept, intensity of light of given wavelength is defined as the number of photons incident per unit area per unit time, with each photon having same energy. Its unit is Wm-2.


6. How will you define threshold frequency?

For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency. The minimum frequency is called threshold frequency.


7. What is a photo cell? Mention the different types of photocells.

Photocell is a device which converts light energy into electrical energy. Its Types are

(i) Photo emissive cell

(ii) Photo voltaic cell

(iii) Photo conductive cell


8. Write the expression for the de Broglie wavelength associated with a charged particle of charge q and mass m, when it is accelerated through a potential V.

A charged particle of charge q and mass m is accelerated through a potential difference of

V. The kinetic energy acquired by the charge is

1/2 mv2 = qV

The speed v of the charge is

 v = √[2qV / m]


The momentum of the charge is

 p = mv = √[2mqV]

The de Broglie wavelength of the charge is

λ =h / mv = h / p

 λ = h/ √[2mqV]



9. State de Broglie hypothesis.

All material particles like electrons, protons, neutrons in motion are associated with waves. These waves are called de Broglie waves or matter waves.


10. Why we do not see the wave properties of a baseball?

The de broglie wavelength

λ = h/p = h/mv

The average speed of the base ball is about 2.24 m/s.

The mass of the ball is 0.625 kg.

 = [6.62 × 10-34  ] / [0.625 × 2.24]

λ = 4.729 x 10-34m

The wavelength of the base ball is almost very very small. So, we cannot see the wave properties of the base ball.


11. A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.

The de Broglie wavelength associated with a charge is 

 λ = h/ √(2mqV)

The charge of an electron and a proton is same.

ie., q = e

The kinetic energy (K) = eV


The kinetic energy of the proton and electron is same. So the de Broglie wavelength depends only the mass of the particle.

The wavelength is inversely proportional to the square root of the mass.

 λ 1/√m

λe / λp = √[mp /me]


The mass of an electron is very lesser than that of proton.

So the electron has greater de Broglie wavelength, than proton. ie., λ e > λ p


12. Write the relationship of de Broglie wavelength λ associated with a particle of mass m in terms of its kinetic energy K.

The de Broglie wavelength of a particle is

 λ = h/mv = h/p

p is the momentum of the particle. The kinetic energy of the particle  (K)  =  ½ mv2  =     p2 /2m

The momentum (p) = √(2mK) 

The de Broglie wavelength of the particle is λ = h / √(2mK)


13. An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?

The de Broglie wavelength of an electron is


The de Broglie wavelength of an alpha particle is


The electron and the alpha particle have same kinetic energy. ie., Ke = Ka

The ratio of λe and λ is

 λe / λ = √[m/me]


The mass of the electron is very lesser than that of the alpha particle. ie., me<<mα

Therefore the de Broglie wavelength of an electron is greater than that of an alpha particle. ie., λe > λ


14. Define stopping potential?

Stopping potential is that the value of the negative potential given to the collecting electrode A which is just sufficient to stop the most energetic photoelectrons emitted and make the photocurrent zero.

The stopping potential is indepentent of intensity of the incident light.

 

15. What is surface barrier?

The potential barrier which prevents the electrons from leaving the metallic surface is called surface barrier. 

 

16. Mention the two features of X-ray spectra, not explained by classical electromagnetic theory.

i) For a given accelerating voltage, the lower limit for the wavelength of continuous X- ray spectra is same for all targets. This minimum wavelength is called cut-off wavelength.

ii) The intensity of X-ray is significantly increased at certain well-defined wavelengths.

 

17. What is Bremsstrahlung?

When a fast moving electron penetrates and approaches a target nucleus, the interaction between the electron and the nucleus either accelerates or decelerates it which results in a change of path of the electron. The radiation produced from such decelerating electron is called Bremsstrahlung or braking radiation.

 

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