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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

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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

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Updated on: Dec 13, 2022

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Physics Part-II (NCERT)

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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.

10 mins ago

Discuss this question LIVE

10 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

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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

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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

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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

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