# 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=A11/3/A21/3 =1 1/3/27 1/3=1/3

4) R= R0A1/3 R1/R2=A11/3 /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

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
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)