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Concepts of Physics

The Nucleus

Find the energy liberated in the reaction { }^{223} {Ra} { }^{

Concepts of Physics

Class 12

HC Verma Part II

Chapter 1: Heat and Temperature

64 questions

Chapter 2: Kinetic Theory of Gases

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Chapter 3: Calorimetry

41 questions

Chapter 4: Laws of Thermodynamics

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Chapter 5: Specific Heat Capacities of Gases

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Chapter 6: Heat Transfer

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Chapter 7: Electric Field and Potential

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Chapter 8: Gauss's Law

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Chapter 10: Electric Current in Conductors

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Chapter 11: Thermal and Chemical Effects of Electric Current

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Chapter 20: Photoelectric Effect and Wave Particle Duality

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Chapter 21: Bohr'a Model and Physics of the Atom

75 questions

Chapter 22: X-rays

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Chapter 23: Semiconductors and Semiconductor devices

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Chapter 24: The Nucleus

94 questions

(a) Calculate the energy released if

^{238} U

emits an

α

-particle.
(b) Calculate the energy to be supplied to

^{238} U

it two protons and two neutrons are to be emitted one by one. The atomic masses of

^{238} U,^{234} Th

and

^{4} He

are

238.0508 u, 234.04363 u

and

4.00260 u

respectively.

(

Use Mass of proton

m_{p} = 1.007276 u

, Mass of

_{1}^{1} H

atom

= 1.007825 u

, Mass of neutron

m_{n} = 1.008665 u

, Mass of electron =

0.0005486 u \approx 511 keV / c^{2}, 1 u = 931 MeV / c^{2} .)

Find the energy liberated in the reaction

^{223} Ra \to^{209} Pb +^{14} C

The atomic masses needed are as follows.

^{223} Ra^{209} Pb^{14} C

22..018 u 208.981 u 14.003 u

Show that the minimum energy needed to separate a proton from a nucleus with Zprotons and N neutrons is

Δ E = (M_{Z - 1, N} + M_{B} - M_{Z, N}) c^{2}

where

M_{Z, N} =

mass of an atom with mass of an atom with Z protons and N neutrons in the nucleus and M

_{B}

= mass of a hydrogen atom. This energy is known as proton-separation energy.

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Question

Easy

Solving time: 3 mins

Find the energy liberated in the reaction
$^{223} Ra \to^{209} Pb +^{14} C$
The atomic masses needed are as follows.
$^{223} Ra^{209} Pb^{14} C$
$22..018 u 208.981 u 14.003 u$

Name: Video solution 1: [Solved] Find the energy liberated in the reaction 223Ra→209 Pb+14C The atomic masses needed are as follows. 223Ra209 Pb14C 22..018u208.981u14.003u - Concepts of Physics
Uploaded: 2022-05-28T09:13:59.000Z
Duration: 4 min
Description: Video solution 1: [Solved] Find the energy liberated in the reaction 223Ra→209 Pb+14C The atomic masses needed are as follows. 223Ra209 Pb14C 22..018u208.981u14.003u - Concepts of Physics

Views: 5,838 students

Updated on: May 28, 2022

Book:

Concepts of Physics (HC Verma Part II)

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Evelyn Discussed

Find the energy liberated in the reaction

^{223} Ra \to^{209} Pb +^{14} C

The atomic masses needed are as follows.

^{223} Ra^{209} Pb^{14} C

22..018 u 208.981 u 14.003 u

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Given:
Atomic mass of

^{223} Ra, m (^{223} Ra) = 223.018 u

Atomic mass of

^{209} Pb, m (^{209} Pb) = 208.981 u

Atomic mass of

^{14} C, m (^{14} C) = 14.003 u

Reaction:

^{223} Ra \to^{209} Pb +^{14} C

Energy,

E = [m (^{223} R a) - (m (^{209} P b) + m (^{14} C))] c^{2} = [223.018 u - (208.981 + 14.003) u] c^{2} = 0.034 \times 931 MeV = 31.65 MeV

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Question Text	Find the energy liberated in the reaction $^{223} Ra \to^{209} Pb +^{14} C$ The atomic masses needed are as follows. $^{223} Ra^{209} Pb^{14} C$ $22..018 u 208.981 u 14.003 u$
Updated On	May 28, 2022
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	294
Avg. Video Duration	3 min

Find the energy liberated in the reaction
$^{223} Ra \to^{209} Pb +^{14} C$
The atomic masses needed are as follows.
$^{223} Ra^{209} Pb^{14} C$
$22..018 u 208.981 u 14.003 u$

Text solutionVerified

Filo tutor solutions (2)

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Find the energy liberated in the reaction 223Ra→209 Pb+14C The atomic masses needed are as follows. 223Ra209 Pb14C 22..018u208.981u14.003u

Text solutionVerified

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Find the energy liberated in the reaction
$^{223} Ra \to^{209} Pb +^{14} C$
The atomic masses needed are as follows.
$^{223} Ra^{209} Pb^{14} C$
$22..018 u 208.981 u 14.003 u$