Browse by Stream
QnA
Home
Search for Exam, Articles, Questions
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. VITEEE- When U(92,235) is bombarded with a neutron, it undergoes as
# Engineering

_{0}n^{1} + _{92}U^{235} \rightarrow _{92}U^{236} \rightarrow _{56}Ba^{144}+ _{36}Kr^{89} + neutrons. In this process, number of neutrons released is

  1. 6

  2. 3

  3. 5

  4. 2

Answers (1)
S Sayak

Total number of neutrons and protons initiaaly = 235 + 1 = 236

Total number of neutrons and protons present in atoms of Ba and Kr = 144 + 89 = 233

Number of neutrons released = 236 - 233 = 3

Similar Questions

Preparation Products

Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions