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Physics Part-II

Nuclei

The Q value of a nuclear reaction, A+b C+d is defined by {Q}=

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 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

Suppose, we think of fission of a

_{26}^{56} F e

nucleus into two equal fragments of

_{13}^{28} A l

. Is the fission energetically possible? Argue by working out Q of the process. Given m

(_{26}^{56} F e)

= 55.93494 u and m

(_{13}^{28} A l)

= 27.98191 u.

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

25 questions

Question

Medium

Solving time: 5 mins

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

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

Views: 6,178 students

Updated on: Aug 24, 2023

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Physics Part-II (NCERT)

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Samuel Discussed

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

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

(i)_{1}^{1} H +_{1}^{3} H \to_{1}^{2} H +_{1}^{2} H Q = [m_{1 H} + m_{1 H}] - [m_{1} + m_{1}_{1}] Q = 1.00758 + 3.016049 - [2.014102 + 2.014102] Q = - 0.0046 u

Since, Q is negative, some mass is converted into energy. Hence, the reaction is endothermic.

(ii)_{6}^{12} C +_{6}^{12} C \to_{10}^{20} Ne +_{2}^{4} He Q = [m_{\frac{1}{6} C} + m_{15 C}] - [m_{11} Ne + m_{\hat{2} He}] Q = 12.000000 + 12.000000 - [19.992439 + 4.00387] Q = 0.00369 u

Since, Q is positive, some mass is converted into energy. Hence, the reaction is exothermic.

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Practice more questions from Physics Part-II (NCERT)

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

Suppose, we think of fission of a

_{26}^{56} F e

nucleus into two equal fragments of

_{13}^{28} A l

. Is the fission energetically possible? Argue by working out Q of the process. Given m

(_{26}^{56} F e)

= 55.93494 u and m

(_{13}^{28} A l)

= 27.98191 u.

View all

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Question Text	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 $m (_{1}^{2} H) = 2.014102 u m (_{1}^{3} H) = 3.016049 u m (_{6}^{12} C) = 12.000000 u m (_{10}^{20} Ne) = 19.992439 u$
Updated On	Aug 24, 2023
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 16
Upvotes	1639
Avg. Video Duration	7 min

Text solutionVerified

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