Home
Class 12
CHEMISTRY
The isotope ""(92)^(235) U decays in a n...

The isotope `""_(92)^(235) U` decays in a number of steps to an isotope of `""_(87)^(207)Pb`. The no of particles emitted in this process will be :

A

`4 alpha`

B

`6 beta`

C

`7 alpha`

D

`4 beta`

Text Solution

Verified by Experts

The correct Answer is:
C, D
`""_(92)U^(235) to ""_(82) U^(2-7) + a (""_(2) alpha^(4)) + b(""_(-1) beta^(0))`
`a = (235- 207)/(4) = 7 alpha , 92- 82 = 2 (7) - b , 10 = 14- b implies b = 4 beta`
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    AAKASH SERIES|Exercise LECTURE SHEET (LEVEL- II) (EXERCISE - II) (Linked Comprehension type questions)|9 Videos

  • NUCLEAR CHEMISTRY

    AAKASH SERIES|Exercise LECTURE SHEET (LEVEL- II) (EXERCISE - III) (Match the following questions )|3 Videos

  • NOBLE GASES

    AAKASH SERIES|Exercise PRACTICE SHEET - 2 (Single more than one option questions)|5 Videos

  • PERIODIC TABLE

    AAKASH SERIES|Exercise QUESTIONS FOR DESCRIPTIVE ANSWERS|37 Videos

Similar Questions

Explore conceptually related problems

The number of neutrons in ""_(92)U^(235) is

The ""_(92)^(235) U disintegrates to give 4 alpha and 6 alpha particles . The atomic number of the end product is

The number of neutrons emitted when ""_(92)^(235) U undergoes controlled nuclear fission to ""_(54)^(142) Xe and ""_(38)^(90) Sr is

Consider the fission of ._(92)^(238)U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primay fragments, are ._(58)^(140)Ce and ._(44)^(99)Ru . Calculate Q for this fission process. The relevant atomic and particle masses are m(._(92)^(238)U)=238.05079 u m(._(58)^(140)Ce)=139.90543 u m(._(44)^(99)Ru)=98.90594 u

The ""_(92)^(238) U disintegrates to give 4 alpha and 6 beta - particles . The atomic number of daughter element obtained is 10 x. x is ___

Which of the following is emitted , when ""_(94)^(239 ) Pu decays into ""_(92) ^(235 ) U ?