Home
Class 12
PHYSICS
The time taken by a photoelectron to com...

The time taken by a photoelectron to come out after the photon strikes is approximately

A

`10^(-4)s`

B

`10^(-10)s`

C

`10^(-16)`

D

`10^(-1) s`

Text Solution

Verified by Experts

The correct Answer is:
B

The photoelectric phenomenon is an instantaneous phenomenon, hence the time taken be an electron to come out of metal is approximately `10^(-10)` sec (found experimentally).
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise-05 [A]|39 Videos
  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise-05 [B]|12 Videos
  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise-04 [A]|20 Videos
  • RACE

    ALLEN|Exercise Basic Maths (Wave Motion & Dopplers Effect) (Stationary waves & doppler effect, beats)|24 Videos
  • TEST PAPER

    ALLEN|Exercise PHYSICS|4 Videos

Similar Questions

Explore conceptually related problems

The work function of a substance is 12.4 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately

Ground state energy of H-atom is (-E_(1)) ,t he velocity of photoelectrons emitted when photon of energy E_(2) strikes stationary Li^(2+) ion in ground state will be:

A particle startsits SHM at t=0 . At a particular instant on its way to extreme position , x= (A)/(2) . Find the time taken by the particle to come back to this point after passing through the extreme position. Hint : t_(1) = time taken to go from x=0 to x= (A)/(2) t_(2) = time taken to go to extreme position and come back to mean position = (T)/(2) Required time = (T)/(2) - t_(1)

The time taken by light to travel from sun to earth is approximately

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy T_(A) eV and De-broglie wavelength lambda_(A) . The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is T_(B) = (T_(A) - 1.50) eV if the de Brogle wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) , then

A photon strikes a hydrogen atom in its ground state to eject the electron with kinetic energy 16.4 eV. If 25% of the photon energy is taken up by the electron, the energy of the incident photon is (24 x X) eV then ‘X’ is:

If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is 10 A , then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximately

A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. if work function of the metal of the foil is 2.2 eV, the time taken by electron to come out is nearly

A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. The experimental observations show that the waiting time for emission is 10^(-8) s. This observation contradicts the calculations based on classical physics view of light energy . Thus we have to assume that during photoelectron emission

Do all the electrons that absorb a photon come out as photoelectrons?

ALLEN-SIMPLE HARMONIC MOTION-Exercise-04 [B]
  1. If the kinetic energy of a free electron doubles . Find the factor by ...

    Text Solution

    |

  2. The threshold frequency for a metallic surface corresponds to an energ...

    Text Solution

    |

  3. The time taken by a photoelectron to come out after the photon strikes...

    Text Solution

    |

  4. The anode voltage of a photocell is kept fixed. The wavelength lamda o...

    Text Solution

    |

  5. Photon of frequency v has a momentum associated with it. If c is the v...

    Text Solution

    |

  6. In an experiment, electrons are made to pass through a narrow slit of ...

    Text Solution

    |

  7. Wave property of electrons implies that they will show diffraction eff...

    Text Solution

    |

  8. Wave property of electrons implies that they will show diffraction eff...

    Text Solution

    |

  9. Wave property of electrons implies that they will show diffraction eff...

    Text Solution

    |

  10. Statement-1 : When ultraviolet light is incidient on a photocell, its ...

    Text Solution

    |

  11. If a source of power 4 kW produces 10^(20) photons/second, the radiati...

    Text Solution

    |

  12. This question has Statement -1 and statement -2. Of the four choices g...

    Text Solution

    |

  13. This question has statement 1 and statement 2. Of the four choices giv...

    Text Solution

    |

  14. The anode voltage of a photocell is kept fixed. The wavelength lamda o...

    Text Solution

    |

  15. If N(0) is the initial mass of a substance with half-life t(1//2) = 5 ...

    Text Solution

    |

  16. At a specific instant emission of radioactive compound is deflected in...

    Text Solution

    |

  17. Which of the following EM ratiation has least wavlenght ?

    Text Solution

    |

  18. A^(238) Unucleus decays by emitting an alpha particle of speed ums^(-1...

    Text Solution

    |

  19. A radioactive sample at any instant has its disintegration rate 5000 d...

    Text Solution

    |

  20. A nucleus with Z =92 emits the following in a sequence: alpha,beta^(...

    Text Solution

    |