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Physics Part-II

Nuclei

The nucleus ^{23}_{11}Ne decays by ^{-} emission. Write down t

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

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

The Q value of a nuclear reaction,

A + b \to C + d

is defined by

Q = [m_{A} + m_{b}] - [m_{C} + m_{d}]

where, the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i)

_{1}^{1} H +_{3}^{1} H \to_{1}^{2} H +_{1}^{2} H

(ii)

_{6}^{12} C +_{6}^{12} C \to_{10}^{20} Ne +_{2}^{4} He

Atomic masses are given to be

m (_{1}^{2} H) = 2.014102 u m (_{1}^{3} H) = 3.016049 u m (_{6}^{12} C) = 12.000000 u m (_{10}^{20} Ne) = 19.992439 u

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

25 questions

Question

Medium

Solving time: 4 mins

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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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

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

The beta decay equation is as shown.

_{10}^{23} N e \to_{11}^{23} N a + e^{-} + Q

Q = Δ m \times

931.5 MeV

Here,

Δ m = m_{N} (_{10}^{23} N e) - m_{N} (_{11}^{23} N a) - m_{e} =

mass defect.

Δ m = [m_{N} (_{10}^{23} N e) - m_{N} (_{11}^{23} N a) - m_{e}] = 22.994466 - 22.089770 = 0.904696

amu

Here, the masses used are masses of nuclei and not of atoms. Note:

m_{e}

has been cancelled.

Q = Δ m \times

931.5 MeV

= 0.904696 \times 931.5 = 4.37

MeV.
The maximum kinetic energy of the electron (max

E_{e}

) = Q = 4.37 MeV.

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Practice more questions from Physics Part-II (NCERT)

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

The Q value of a nuclear reaction,

A + b \to C + d

is defined by

Q = [m_{A} + m_{b}] - [m_{C} + m_{d}]

where, the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i)

_{1}^{1} H +_{3}^{1} H \to_{1}^{2} H +_{1}^{2} H

(ii)

_{6}^{12} C +_{6}^{12} C \to_{10}^{20} Ne +_{2}^{4} He

Atomic masses are given to be

m (_{1}^{2} H) = 2.014102 u m (_{1}^{3} H) = 3.016049 u m (_{6}^{12} C) = 12.000000 u m (_{10}^{20} Ne) = 19.992439 u

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Question Text	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: $m (_{10}^{23} N e) = 22.994466 u$ $m (_{11}^{23} N a) = 22.089770 u$ .
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1

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:
$m (_{10}^{23} N e) = 22.994466 u$
$m (_{11}^{23} N a) = 22.089770 u$ .

Text solutionVerified

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The nucleus 1123​Ne decays by β− emission. Write down the β - decay equation and detere the maximum kinetic energy of the electrons emitted. Given that:m(1023​Ne)=22.994466um(1123​Na)=22.089770u.

Text solutionVerified

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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:
$m (_{10}^{23} N e) = 22.994466 u$
$m (_{11}^{23} N a) = 22.089770 u$ .