Home
Class 12
CHEMISTRY
Radioactive decay follows first-order ki...

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are `tau = 1// lambda` and `t_(1//2) = 0.693//lambda`. Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after `n` half lives of radioactive elements can be calculated using the relation:
`N = N_(0) ((1)/(2))^(n)`
The rate of radioactive decay is

A

Independent of time

B

Independent of temperature

C

Dependent on catalyst

D

Dependent on the amount of elementsd not yet decayed

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Multiple Correct|26 Videos

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Single Correct|80 Videos

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex6.5 Objective|10 Videos

  • NCERT BASED EXERCISE

    CENGAGE CHEMISTRY ENGLISH|Exercise Nuclear Chemistry (NCERT Exercise)|29 Videos

  • ORGANIC COMPOUNDS WITH FUNCTIONAL GROUP

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Analytical And Descriptive|24 Videos

Similar Questions

Explore conceptually related problems

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Select the correct relation.

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Amount of radioactive elements (activity) decreases with passage of time as

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Which "is"//"are" true about the decay cosntant?

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Half life of .^(60)Co is 5.3 years, the time taken for 99.9% decay will be

A fraction f_1 of a radioactive sample decays in one mean life, and a fraction f_2 decays in one half life. Then

n alpha particles per second are emitted from N atoms of a radioactive element. Then find the half life of radioactive element ?

N atoms of a radioactive element emit n alpha particles per second. The half-life of tge element is.

The radioactive decay rate of a radioactive element is found to be 10^(3) disintegration per second at a certain time . If the half life of the element is one second , the decay rate after one second is?

In a sample of a radioactive substance, what fraction of the initial nuceli will remain undecayed after a time t=T//2 , where T=half -life of radioactive substance?

If T is the half-life of a radioactive material, then the fraction that would remain after a time (T)/(2) is

CENGAGE CHEMISTRY ENGLISH-NUCLEAR CHEMISTRY-Exercises Link Comprehension
  1. Unstable nuclei attain stability through disintegration. The nuclear s...

    Text Solution

    |

  2. Unstable nuclei attain stability through disintegration. The nuclear s...

    Text Solution

    |

  3. Unstable nuclei attain stability through disintegration. The nuclear s...

    Text Solution

    |

  4. In the disintegration of a radioactive element, alpha- and beta-partic...

    Text Solution

    |

  5. In the disintegration of a radioactive element, alpha- and beta-partic...

    Text Solution

    |

  6. In the disintegration of a radioactive element, alpha- and beta-partic...

    Text Solution

    |

  7. In the disintegration of a radioactive element, alpha- and beta-partic...

    Text Solution

    |

  8. In the decay series .(92)U^(238) to .(82)Pb^(206), how many alpha-pari...

    Text Solution

    |

  9. Uranium .(92)U^(238) decayed to .(82)Pb^(206). They decay process is ....

    Text Solution

    |

  10. Uranium .(92)U^(238) decayed to .(82)Pb^(206). They decay process is ....

    Text Solution

    |

  11. Uranium .(92)U^(238) decayed to .(82)Pb^(206). They decay process is ....

    Text Solution

    |

  12. The source of energy of stars is nuclear fusion. Fusion reaction occur...

    Text Solution

    |

  13. The source of energy of stars is nuclear fusion. Fusion reaction occur...

    Text Solution

    |

  14. The source of energy of stars is nuclear fusion. Fusion reaction occur...

    Text Solution

    |

  15. The source of energy of stars is nuclear fusion. Fusion reaction occur...

    Text Solution

    |

  16. Radioactive decay follows first-order kinetic. The mean life and half-...

    Text Solution

    |

  17. Radioactive decay follows first-order kinetic. The mean life and half-...

    Text Solution

    |

  18. Radioactive decay follows first-order kinetic. The mean life and half-...

    Text Solution

    |

  19. Radioactive decay follows first-order kinetic. The mean life and half-...

    Text Solution

    |

  20. Radioactive decay follows first-order kinetic. The mean life and half-...

    Text Solution

    |