DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

JEE: DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

The document DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

Introductory Exercise 31.1

Ques 1: Activity of a radioactive substance decreases from 8000 8q to 1000 Bq in 9 days. What is the half life and average life of the radioactive substance?
Sol:
 DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
∴ n = 3 = number of half lives There half lives are equivalent to 9 days. Hence one half life is 3 days.
tav = 1.44 t1/2 = 1.44 × 3 = 4.32 days

Ques 2: A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity  (t = 0) of 40 μCi. Calculate the number of nuclei that decay in the time interval between t1 = 10.0 hand t2 =12.0 h.
Sol: 
R= λN0
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
where, DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
N = Noe-λt
Find N1 = Noe-λt
and N2 =  Noe-λt
Number of nuclei decayed in given time= N1 N2

Ques 3: A freshly prepared sample of a certain radioactive isotope has an activity of 10 mCi. After 4.0 h its activity is 8.00 mCi.
(a) Find the decay constant and half life
(b) How many atoms of the isotope were contained in the freshly prepared sample?
(c) What is the sample's activity 30.0 h after it is prepared?
Sol: 
(a)R = R0e-λt
R0 - 20 mci
R = 8 mci
t = 4.0 h Find λ.
(b) R0 =λN0
Find. DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
(c) Find R = R0e-λt

Ques 4: A radioactive substance contains 1015 atoms and has an activity of 6.0 × 1011 Bq. What is its half-life?
Sol: 
R0 = λN0
6.0 × 1011 = λ(1015)
∴ λ = 6.0 × 10-4 s
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 5: Two radioactive elements X and Y have half-life periods of 50 minutes and 100 minutes respectively. Initially both of them contain equal number of atoms. Find the ratio of atoms left NX/NY after 200 minutes.
Sol:
In 200 minute time,
n1 = number of half lives of X
= 200/50 = 4
n2 = number of half lives of Y
= 200/100 = 2
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE 

Introductory Exercise 31.2

Ques 1: (a) How much mass is lost per day by a nuclear reactor operated at a 109 watt power level?
(b) If each fission releases 200 MeV, how many fissions occur per second to yield this power level?
Sol: 
(a)m(c)2 =P x t
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
(b) Number of fissions required per second
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 3.125 × 1019

Ques 2: Find energy released in the alpha decay
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
Given
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
Sol:
Mass defect ∑mi - ∑m= Δm
= (238.050784) - (234.043593 + 4.002602) = 4.589 × 10-3 u
Energy released = Δm × 931.48 MeV = 4.27 MeV

Exercises
For JEE Main

Subjective Questions
Radioactivity

Ques 1: The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. Five minutes later the rate becomes 2700 per minute. Calculate
(a) decay constant and
(b) half-life of the sample
Sol: (a) R = R0e-λt 
R = 2700 per minute, R0 = 4750 per minute t = 5 minute
Find λ.
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 2: A radioactive sample contains 1.00 × 1015 atoms and has an activity of 6.00 × 1011 Bq. What is its half-life?
Sol: 
R = λN
6 × 1011 =1.0 × l015λ
λ= 6 × 10-4s
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE = 1155 s= 19.25 min

Ques 3: Obtain the amount of 60 Co necessary to provide a radioactive source of 8.0 Ci strength. The hal-flife of 60Co is 5.3 years?
Sol:
R = λN
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 7.14 × 1019
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 4: The half-life of DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE gainst alpha decay is 4.5 × 109 year. How much disintegration per second occurs in 1 g of DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE?
Sol:
 DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 1.23 x 104dps 5. 1/λ = 10days
∴ λ = 0.1 day-1
Probability of decay
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 1 – e-λt = 1 - e-0.1×5 = 0.39

Ques 5: In an ore containing Uranium, the ratio of 238U to 206Pb nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of 238U. Take the half-life of 238U to be 4.5 × 10years.
Sol: 
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
N0 = 3 + 1 = 4 N =3
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
From Eqs. (i) and (ii), we get t = 1.88 × 109yr

Ques 6: The half-lives of radioisotopes P32 and P33 are 14 days and 25 days respectively. These radioisotopes are mixed in the ratio of 4 :1 of their atoms. If the initial activity of the mixed sample is 3.0 m Ci, find the activity of the mixed isotopes after 60 year.
Sol:
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 7: Consider two decay reactions.
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE Pb+ 10 protons + 20 neutrons
DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE
Are both the reactions possible?
Sol: 
(a) 82 + 10 = 92, 206 + 10 + 20 = 236
So this reaction is possible.
(b) 82 + 16 - 6 = 92, 206 + 32 = 238
But antineutrino is also emitted with β~1 (or electron) decay.

Ques 8: Obtain the binding energy of a nitrogen nucleus from the following data :
mH = 1.00783 u, mN = 1.00867 u, m(147 N) = 14.00307 u
Give your answer in units of MeV. [Remember 1 u = 931.5 Me V/c2]
Sol:
Binding energy = Δm × 931.5 MeV
= (7 × 1.00783 + 7 × 1.00867-14.00307) 931.5 = 104.72 MeV

The document DC Pandey Solutions: Modern Physics II Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

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