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Physics Part-II

Nuclei

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

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

Obtain approximately the ratio of the nuclear radii of the gold isotope

_{79}^{197} A u

and the silver isotope

_{47}^{107} A g

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.

View all question

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

25 questions

Question

Medium

Solving time: 5 mins

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:
$m (_{88}^{226} R a) = 226.02540 u$ ,
$m (_{86}^{222} R n) = 222.01750 u$ ,
$m (_{84}^{216} P o) = 216.00189 u$

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Physics Part-II (NCERT)

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Charlotte Discussed

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

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

(a)

_{88}^{226} R a \to_{86}^{222} R n +_{2}^{4} α

_{86}^{222} R n \to_{84}^{216} P o +_{2}^{4} α

Q = Δ m \times 931 M e V

Δ m = 226.02540 - 222.01750 - 4.002603 = 0.005297

Q = 4.93 MeV,

E_{α} = 4.85 M e V

Kinetic energy

= \frac{M a ss n o . a f t er d ec a y}{M a ss n o . b e f ore d ec a y} \times Q

K . E = \frac{222}{226} \times 4.93 = 4.85 M e V

(b)

_{86}^{222} R n \to_{84}^{216} P o +_{2}^{4} α

Q = Δ m \times 931 M e V

Δ m = 220.01137 - 216.00189 - 4.002603 = 0.006877

Q = 6.4 MeV,

E_{α} = 6.29 M e V

Kinetic energy

= \frac{M a ss n o . a f t er d ec a y}{M a ss n o . b e f ore d ec a y} \times Q

K . E = \frac{216}{222} \times 6.4 = 6.29 M e V

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Practice more questions from Physics Part-II (NCERT)

Obtain approximately the ratio of the nuclear radii of the gold isotope

_{79}^{197} A u

and the silver isotope

_{47}^{107} A g

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.

View all

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Question Text	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: $m (_{88}^{226} R a) = 226.02540 u$ , $m (_{86}^{222} R n) = 222.01750 u$ , $m (_{84}^{216} P o) = 216.00189 u$
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	32

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:
$m (_{88}^{226} R a) = 226.02540 u$ ,
$m (_{86}^{222} R n) = 222.01750 u$ ,
$m (_{84}^{216} P o) = 216.00189 u$

Text solutionVerified

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Find the Q-value and the kinetic energy of the emitted α- Particle in the α-decay of (a) 86220​Rn and (b) 88226​Ra.Given:m(88226​Ra)=226.02540u,m(86222​Rn)=222.01750u,m(84216​Po)=216.00189u

Text solutionVerified

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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:
$m (_{88}^{226} R a) = 226.02540 u$ ,
$m (_{86}^{222} R n) = 222.01750 u$ ,
$m (_{84}^{216} P o) = 216.00189 u$