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Chemistry Homework Help => CBSE XI and XII Chemistry => Topic started by: Pranjal Singh on August 13, 2020, 05:51:57 AM

Title: Nuclear Chemistry
Post by: Pranjal Singh on August 13, 2020, 05:51:57 AM
For the given series reaction in nth step, find out the number of produced neutrons and energy-
238U→Ba+Kr+3n+Energy(E)
Title: Re: Nuclear Chemistry
Post by: uma on August 15, 2020, 05:23:49 AM
Every step releases three neutrons during this decay series.
And it is a chain reaction in which every neutron is going to create a new step to generate 3 more neutrons
It means up to nth step 3n  neutrons are released
Energy for the first step is E .hence E is 3n-1 E