Search Colleges, Courses, Exams, QnA, & Articles

LoginSign Up
Physics Ncert Solutions Class 12thNucleiPhysics NCERT Exemplar Solutions Class 12th Chapter ThirteenClass 12thPhysics

He23 and He13 nuclei have the same mass number. Do they have the same binding energy?

1 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
A
8 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- nuclei 2He4 and 1He3 have the same mass number . the element having more number of proton having more repusion so less binding energy. So 2He4 have more number of proton so more repulsion so less binding energy.

Thumbs Up IconUpvote Thumbs Down Icon

Similar Questions for you

The distance of closest approach of an alpha particle fired towards gold nucleus with kinetic energy k is r0. The distance of closest approach when the alpha particle is fired towards same nucleus with kinetic energy 4k will be

AK
Alok Kumar Singh
Kindly go through the solution

 

Find the mass number of an atom whose radius is half of that of a given atom of mass number 192.

AK
Alok Kumar Singh

r = R 0 ( 1 9 2 ) 1 3

r 2 = R 0 ( m ) 1 3

m = 1 9 2 8 = 2 4

A nuclear fusion reaction is given by   If atomic mass of He is 4.0026 amu and that of O is 15.9994 amu then the Q-value of the reaction will be approximately

AK
Alok Kumar Singh

Q = [4 *4.0026 – 15.9994] *931.5 MeV

Q = 10.2 MeV

A radioactive simple is undergoing a decay. At any time t1, its activity is A and another time t2, the activity is A 5 . <!-- [if gte mso 9]><xml> </xml><![endif]--> What is the average life time for the sample?

AK
Alok Kumar Singh

A = A 0 e − λ t 1     [Radio active decay law]

A 5 = A 0 e − λ ( t 2 − t 1 )

⇒ A v e r a g e     l i f e = 1 λ = ( t 2 − t 1 ) l n 5  

          

...Read more

A sample contains 10-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is x 1 0 0 . The value of x is ______.

AK
Alok Kumar Singh

f o r     A , t 1 2 = 4     s e c  

= m A 0 e − 0 . 6 9 3 4 × 1 6

m A = m A 0 e − 4 × 0 . 6 9 3    -(1)

for B,

for B,  t 1 2 = 8 s e c  

m B = m B 0 e − 2 × 0 . 6 9 3               -(2)

m A m B = m A O m B O e − 4 × 0 . 6 9 3 e − 2 × 0 . 6 9 3

m A m B = 2 5 1 0 0 = x 1 0 0 x = 2 5

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System
QuestionsDiscussionsTags

Have a question related to your career & education?

or

See what others like you are asking & answering