Home
Class 12
PHYSICS
Crystal diffraction experiments can be p...

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to `1 Å`, which is of the order of inter-atomic spacing in the lattice) `(m_(e) =9.11 × 10^(-31) kg).`

Text Solution

Verified by Experts

For `lambda=1Å` electron’s energy `= 150 eV`, photon.s energy `= 12.4 keV`. Thus, for the same wavelength, a photon has much greater energy than an electron.
Promotional Banner

Topper's Solved these Questions

  • DUAL NATURE OF RADIATION AND MATTER

    NCERT BANGLISH|Exercise EXERCISES|41 Videos

  • CURRENT ELECTRICITY

    NCERT BANGLISH|Exercise ADDITIONAL EXERCISES|5 Videos

  • ELECTRIC CHARGES AND FIELDS

    NCERT BANGLISH|Exercise EXERCISES|29 Videos

Similar Questions

Explore conceptually related problems

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second from a light source of power 40 W and wavelength 5893 Å

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . Two stationary electrons are accelerated with potential difference V_(1) and V_(2) respectively such that V_(1) : V_(2) = n . The ratio of their de Broglie wavelength

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second by a source of light of power 30 W is 10^(20) , the momentum of each photon (in kg.m.s^(-1) )

A moving electron has 4.9 xx 10^(-25) joules of kinetic energy. Find out its de - Broglie wavelength (Given h = 6.626 xx 10^(-34) Js, m_e = 9.1 xx 10^(-31) kg).

Coulomb’s law for electrostatic force between two point charges and Newton’s Law of gravitational force between masses depends on the distance between the charges/ masses. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1overset@A(=10^(-10)m) apart? (m_p=1.67x10^(-27)kg,m_e=9.11x10^(-31)kg)

Calculate the kinetic energy of a moving electron which has a wavelength of 4.8 pm. [mass of electron = 9.11 xx 10^(-31) kg, h = 6.626 xx 10^(-34) Kg m^2s^(-1) ].

NCERT BANGLISH-DUAL NATURE OF RADIATION AND MATTER-EXERCISES
  1. An electron gun with its collector at a potential of 100 v fires out e...

    Text Solution

    |

  2. (a) An X-ray tube produces a continuous spectrum of radiation with its...

    Text Solution

    |

  3. In an accelerator experiment on high-energy collisions of electrons wi...

    Text Solution

    |

  4. Estimating the following two numbers should be interesting. The first ...

    Text Solution

    |

  5. Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irr...

    Text Solution

    |

  6. Monochromatic radiation of wavelength "640.2 nm "(1nm = 10^(-9) m) fro...

    Text Solution

    |

  7. A mercury lamp is a convenient source for studying frequency dependenc...

    Text Solution

    |

  8. The work function for the following metals is given: Na: 2.75 eV, K: 2...

    Text Solution

    |

  9. Light of intensity 10^(-5) W m^(-2) falls on a sodium photo-cell of su...

    Text Solution

    |

  10. Crystal diffraction experiments can be performed using X-rays, or elec...

    Text Solution

    |

  11. (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 15...

    Text Solution

    |

  12. An electron microscope uses electrons accelerated by a voltage of 50 k...

    Text Solution

    |

  13. The wavelength of a probe is roughly a measure of the size of a struct...

    Text Solution

    |

  14. Find the typical de Broglie wavelength associated with a He atom in he...

    Text Solution

    |

  15. Compute the typical de Broglie wavelength of an electron in a metal at...

    Text Solution

    |

  16. Answer the following questions: (a) Quarks inside protons and neutro...

    Text Solution

    |

  17. Answer the following questions: (b) What is so special about the com...

    Text Solution

    |

  18. Answer the following questions: (c) Why should gases be insulators a...

    Text Solution

    |

  19. Answer the following questions: (d) Every metal has a definite work ...

    Text Solution

    |

  20. Answer the following questions: (e) The energy and momentum of an el...

    Text Solution

    |