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CBSE Physics (Chapter Wise With Hint / Solution) Class XII

CBSE Guess > eBooks > Class XII > CBSE Physics (Chapter Wise With Hint / Solution) Class XII

PHYSICS - CBSE CLASS XIIPHYSICS - CBSE CLASS XII

Atomic Nucleus Key

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1) The entire positive charge and the mass were concentrated at one place inside the

atom, called the nucleus.

2) A larger number of alpha particle went through undeflected

3) R=R0A1/3 R1 /R2=A1

1/3 /A21/3 =1 1/3 /27 1/3=1/3

4) R= R0A1/3 R1 /R2=A1

1/3 /A21/3=(1/8)1/3 =1/2

R1: R2=1:2

5) α-particles have more ionizing power than β-particles.

6) N/N0=(1/2)n

t=2T =2*30 =60 days

7) They are neutral in nature and get absorbed by nucleus, thus distributing the neutron proton ratio.

8) The ratio of neutrons to proton ratio increases, after the emission of a ά – particle.

9) Owing to greater mass and charge, it is able to knock out/pull out electrons which colliding with atoms and molecules in its path.

10) T = 20 minutes t = 60 minutes N/No = (1/2)n = (1/2)60/20 = (1/2) 3 =1/8 After one hour, 1/8th of the original mass would remain.

11) The nucleus looses energy, but remains same isotope it was.

12) No a nucleus either emits a ά – particle or a β – particle and if left in the exited state, it may emit γ – ray also.

13) 1:1 (independent of A).

14) At t = T ½

N = No / 2 Using N = No e - λt

No / 2 = No e- λt½

Solving we get, T1/2 = ln2/ λ = 0.693/ λ

15) Size of nucleus can approximately be estimated using the concept of distance of closest approach. The rebounding particle is selected and

its information is substituted in the expression Ro = 1/4πEo x 2Ze2 / E for ά particle where E is its energy.

B. Using N =No (1/2)t/ π /2 N =1/16No 1/16 = ½(1/2)t/ π /2 T1/2 = 30/4 = 7.5 days.

17) 235 = 142+Y+3 Y = 90

And 92 = 57+Z+0

Z = 35

18) a) Yes. Since X & Y are having same atomic number.

b) Y43 is likely to be more stable because for its neutron to proton ratio is smaller.

19) Disintegration constant λ =0.693/T1/2

=0.693/30x24x60x60

Therefore T avg =1.44xT1/2 = 1.44x30 = 43.2 days.

20) Remaining amount (undecayed) =1/4N0

Using N= N0 (1/2)t/T1/2

1/4= (1/2)t/60

Solving t=2x60=120days

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Atomic Nucleus Solution

Current Electricity Solution

Communications Systems Solution

Electronic Devices Solution

Electrostatics Solution

Optics Solution

Electromagnetic Wave Solution

Dual Nature Of Matter And Rediations Solution

Magnetic Effects Of Current & Magnetism Solution

Electromagnetic Induction & Alternating Current Solution

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21) Mass defect Δm= (22.9945-22.9898) =0.00474

Energy Q= (0.00474) (931.5)

=4.4MeV

Hence the energy of beta particle can range from 0 to 4.4MeV.

22) a) Using R=R0e-λt

2700=4750 e-5t

λ=0.113min-1

b) Using T1/2=0.693/ λ =0.693/0.113 =6.132min

23) In the process of beta decay, a neutron gets converted to proton inside the nucleus . Hence number of neutrons decreases by one whereas

number of proton increases by one. Hence n/p ratio decreases.

210Bi83--------------------210Po84 +

0β-1 + γ

Before decay =127/83

After decay=126/84

24) Let λ and λ’ be the decay constant of element A and B respectively. Given is

T1/2(A) = T1/2(B)

0.693/ λ=1/ λ’ or λ / λ’ =0.693

Let N be the number of atoms of each of the two samples and R and R’ their disintegration rate, then

R/R’= λN/ λ’N= λ / λ’=0.693

→R’ > R

25) Find Δm using

Δm= (7x1.00783+7x1.00867-14.003074) U

Calculate ΔEb=Δmx931.5 MeV

26) a) 226Ra88--------------→222Rn86 +4He2

b) 32P15------------------→32S16+

0e-1 +γ

c) 32P15------------------→11B5 +

0e+1 + γ

27) i) 6Li3 +1n0 ----------→3H1 +

4He2 +Q (energy)

ii) Q=Δmx931 MeV

Where Δm=6.01512+1.0086654-4.0026044-3.0100000

 

CBSE Physics (Chapter Wise With Hint / Solution) Class XII (By Mr. Sreekumaran Nair)

Physics - Mr. Sreekumaran Nair

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