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Physics Part-II

Nuclei

A source contains two phosphorous radio nuclides ^{32}_{15}P,

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 neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energy of the nuclei

_{13}^{27} A l

from the following data:

m (_{13}^{26} A l) = 25.986895 u

m (_{13}^{27} A l) = 26.981541 u

A source contains two phosphorous radio nuclides

_{15}^{32} P (T_{1/2} = 14.3 d)

and

_{15}^{33} P (T_{1/2} = 25.3 d)

. Initially, 10% of the decays come from

_{15}^{32} P

. How long one must wait until 90% to do so?

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an

α

-particle. Consider the following decay processes:

_{88}^{223} R a \to_{82}^{209} P b +_{6}^{14} C

_{88}^{223} R a \to_{86}^{219} R n +_{2}^{4} He

Calculate the Q-values for these decays and detere that both are energetically allowed.

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

25 questions

Question

Medium

Solving time: 5 mins

A source contains two phosphorous radio nuclides $_{15}^{32} P (T_{1/2} = 14.3 d)$ and $_{15}^{33} P (T_{1/2} = 25.3 d)$ . Initially, 10% of the decays come from $_{15}^{32} P$ . How long one must wait until 90% to do so?

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Physics Part-II (NCERT)

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A source contains two phosphorous radio nuclides

_{15}^{32} P (T_{1/2} = 14.3 d)

and

_{15}^{33} P (T_{1/2} = 25.3 d)

. Initially, 10% of the decays come from

_{15}^{32} P

. How long one must wait until 90% to do so?

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Let the number of atoms of

_{15}^{32} P

N_{1}

and atoms of

_{15}^{33} P

N_{2}

The decay constants of two atoms be

λ_{1}

and

λ_{2}

respectively.

The initial activity of

_{15}^{33} P

\frac{1}{9}

times that of

_{15}^{32} P

Therefore, we can write the decay constants as:

N_{1} λ_{1} = \frac{N _{2} λ _{2}}{9}

___(i)

Let after time t the activity of

_{15}^{33} P

9

times that of

_{15}^{32} P

N_{1} λ_{1} e^{- λ_{1} t} = 9 N_{2} λ_{2} e^{- λ_{2} t}

___(ii)

Now. Divide equation (ii) by (i) and taking the natural log of both sides we get

- λ_{1} t = l n 81 - λ_{2} t

t = \frac{l n 81}{λ _{2} - λ _{1}}

where

λ_{2} = 0.048/ d a y

and

λ_{1} = 0.027/ d a y

t = \frac{4.394}{0.048 - 0.027} = 209.2 d a ys

t comes out to be

209.2

days

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Practice more questions from Physics Part-II (NCERT)

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energy of the nuclei

_{13}^{27} A l

from the following data:

m (_{13}^{26} A l) = 25.986895 u

m (_{13}^{27} A l) = 26.981541 u

A source contains two phosphorous radio nuclides

_{15}^{32} P (T_{1/2} = 14.3 d)

and

_{15}^{33} P (T_{1/2} = 25.3 d)

. Initially, 10% of the decays come from

_{15}^{32} P

. How long one must wait until 90% to do so?

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an

α

-particle. Consider the following decay processes:

_{88}^{223} R a \to_{82}^{209} P b +_{6}^{14} C

_{88}^{223} R a \to_{86}^{219} R n +_{2}^{4} He

Calculate the Q-values for these decays and detere that both are energetically allowed.

View all

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Question Text	A source contains two phosphorous radio nuclides $_{15}^{32} P (T_{1/2} = 14.3 d)$ and $_{15}^{33} P (T_{1/2} = 25.3 d)$ . Initially, 10% of the decays come from $_{15}^{32} P$ . How long one must wait until 90% to do so?
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	24

A source contains two phosphorous radio nuclides $_{15}^{32} P (T_{1/2} = 14.3 d)$ and $_{15}^{33} P (T_{1/2} = 25.3 d)$ . Initially, 10% of the decays come from $_{15}^{32} P$ . How long one must wait until 90% to do so?

Text solutionVerified

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A source contains two phosphorous radio nuclides $_{15}^{32} P (T_{1/2} = 14.3 d)$ and $_{15}^{33} P (T_{1/2} = 25.3 d)$ . Initially, 10% of the decays come from $_{15}^{32} P$ . How long one must wait until 90% to do so?