Get 2 FREE Instant-Explanations on Filo with code FILOAPP

Home

Physics Part-II

Nuclei

The radio nuclide ^{11}C decays according to ^{11}_{6}C, , ^{1

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

Chapter 2: Wave Optics

30 questions

Chapter 3: Dual Nature of Radiation and Matter

44 questions

Chapter 4: Atoms

23 questions

Chapter 5: Nuclei

38 questions

Find the Q-value and the kinetic energy of the emitted

α

- Particle in the

α

-decay of (a)

_{86}^{220} R n

and (b)

_{88}^{226} R a

Given:

m (_{88}^{226} R a) = 226.02540 u

m (_{86}^{222} R n) = 222.01750 u

m (_{84}^{216} P o) = 216.00189 u

The radio nuclide

^{11} C

decays according to

_{6}^{11} C \to_{5}^{11} B + e^{+} + ν

T_{1/2}

= 20.3

The maximum energy of the emitted positron is

0.960 M e V

.
Given the mass values:

m (_{6}^{11} C) = 11.011434 u

and

m (_{6}^{11} B) = 11.009305 u

Calculate

Q

and compare it with the maximum energy of the positron emitted.

The nucleus

_{11}^{23} N e

decays by

β^{-}

emission. Write down the

β

- decay equation and detere the maximum kinetic energy of the electrons emitted. Given that:

m (_{10}^{23} N e) = 22.994466 u

m (_{11}^{23} N a) = 22.089770 u

View all question

Chapter 6: Semiconductor Electronics: Materials, Devices and Simple Circuits

25 questions

Question

Medium

Solving time: 4 mins

The radio nuclide $^{11} C$ decays according to
$_{6}^{11} C \to_{5}^{11} B + e^{+} + ν$ : $T_{1/2}$ $= 20.3$
The maximum energy of the emitted positron is $0.960 M e V$ .
Given the mass values:
$m (_{6}^{11} C) = 11.011434 u$ and $m (_{6}^{11} B) = 11.009305 u$
Calculate $Q$ and compare it with the maximum energy of the positron emitted.

Name: Video solution 1: [Solved] The radio nuclide 11C decays according to611C→511B+e++ν: T1/2 =20.3 The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:m(611C)=11.011434u and m(611B)=11.009305uCalculate Q and compare it with the maximum energy of the positron emitted. - Physics Part-II
Uploaded: 2022-12-13T14:06:55.000Z
Duration: 8 min
Description: Video solution 1: [Solved] The radio nuclide 11C decays according to611C→511B+e++ν: T1/2 =20.3 The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:m(611C)=11.011434u and m(611B)=11.009305uCalculate Q and compare it with the maximum energy of the positron emitted. - Physics Part-II

Views: 6,368 students

Updated on: Dec 13, 2022

Book:

Physics Part-II (NCERT)

Found 7 tutors discussing this question

Sofia Discussed

The radio nuclide

^{11} C

decays according to

_{6}^{11} C \to_{5}^{11} B + e^{+} + ν

T_{1/2}

= 20.3

The maximum energy of the emitted positron is

0.960 M e V

.
Given the mass values:

m (_{6}^{11} C) = 11.011434 u

and

m (_{6}^{11} B) = 11.009305 u

Calculate

Q

and compare it with the maximum energy of the positron emitted.

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

_{6}^{11} C \to_{6}^{11} B + e^{+} + ν + Q

Q = [m_{N} (_{6}^{11} C) - m_{N} (_{6}^{11} B) - m_{e}] c^{2}

Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add

6 m_{e}

in case of

^{11} C

and

5 m_{e}

in case of

^{11} B

. Hence,

Q = [m (_{6}^{11} C) - m (_{6}^{11} B) - 2 m_{e}] c^{2}

(Note

m_{e}

has been doubled)
Using given masses, Q = 0.961 MeV.
Q =

E_{d} + E_{e} + E_{ν}

The daughter nucleus is too heavy compared to

e^{+}

and

ν

, so it carries negligible energy (

E_{d}

\approx

0). If the kinetic energy (

E_{ν}

) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum

E^{e}

\approx

Was this solution helpful?

Report

Filo tutor solutions (1)

Learn from their 1-to-1 discussion with Filo tutors.

8 mins

Uploaded on: 12/13/2022

Ask your question, on a video call with tutor

Connect instantly with this tutor

Connect now

Taught by

Harsh Kumar Verma

Total classes on Filo by this tutor - 444

Teaches : Physics, Organic Chemistry, Inorganic Chemistry

Connect instantly with this tutor

Connect now

Was this solution helpful?

Report

One destination for complete JEE/NEET preparation

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Practice more questions from Physics Part-II (NCERT)

Find the Q-value and the kinetic energy of the emitted

α

- Particle in the

α

-decay of (a)

_{86}^{220} R n

and (b)

_{88}^{226} R a

Given:

m (_{88}^{226} R a) = 226.02540 u

m (_{86}^{222} R n) = 222.01750 u

m (_{84}^{216} P o) = 216.00189 u

The radio nuclide

^{11} C

decays according to

_{6}^{11} C \to_{5}^{11} B + e^{+} + ν

T_{1/2}

= 20.3

The maximum energy of the emitted positron is

0.960 M e V

.
Given the mass values:

m (_{6}^{11} C) = 11.011434 u

and

m (_{6}^{11} B) = 11.009305 u

Calculate

Q

and compare it with the maximum energy of the positron emitted.

The nucleus

_{11}^{23} N e

decays by

β^{-}

emission. Write down the

β

- decay equation and detere the maximum kinetic energy of the electrons emitted. Given that:

m (_{10}^{23} N e) = 22.994466 u

m (_{11}^{23} N a) = 22.089770 u

View all

Stuck on the question or explanation?

Connect with our Physics tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	The radio nuclide $^{11} C$ decays according to $_{6}^{11} C \to_{5}^{11} B + e^{+} + ν$ : $T_{1/2}$ $= 20.3$ The maximum energy of the emitted positron is $0.960 M e V$ . Given the mass values: $m (_{6}^{11} C) = 11.011434 u$ and $m (_{6}^{11} B) = 11.009305 u$ Calculate $Q$ and compare it with the maximum energy of the positron emitted.
Updated On	Dec 13, 2022
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	144
Avg. Video Duration	8 min

Text solutionVerified

Filo tutor solutions (1)

Practice more questions from Physics Part-II (NCERT)

Practice questions from Physics Part-II (NCERT)

Practice more questions from Nuclei

Practice questions on similar concepts asked by Filo students

The radio nuclide 11C decays according to611​C→511​B+e++ν: T1/2​ =20.3 The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:m(611​C)=11.011434u and m(611​B)=11.009305uCalculate Q and compare it with the maximum energy of the positron emitted.

Text solutionVerified

Filo tutor solutions (1)

Practice more questions from Physics Part-II (NCERT)

Practice questions from Physics Part-II (NCERT)

Practice more questions from Nuclei

Practice questions on similar concepts asked by Filo students

Related books for questions on Nuclei