Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

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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:
 Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
∴ 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
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
where, Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
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. Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
(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
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

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
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev 

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
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Number of fissions required per second
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
= 3.125 × 1019

Ques 2: Find energy released in the alpha decay
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
Given
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
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.
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

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
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev = 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
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
= 7.14 × 1019
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 4: The half-life of Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev gainst alpha decay is 4.5 × 109 year. How much disintegration per second occurs in 1 g of Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev?
Sol:
 Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
= 1.23 x 104dps 5. 1/λ = 10days
∴ λ = 0.1 day-1
Probability of decay
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
= 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: 
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
N0 = 3 + 1 = 4 N =3
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
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:
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 7: Consider two decay reactions.
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev Pb+ 10 protons + 20 neutrons
Modern Physics II- Physics, Solution by DC Pandey NEET Notes | EduRev
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

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