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=R_{0}A^{1/3} R_{1}/R_{2}=A_{1}^{1/3}/A_{2}^{1/3} =1 ^{1/3}/27 ^{1/3}=1/3

4) R= R_{0}A^{1/3} R_{1}/R_{2}=A_{1}^{1/3} /A_{2}^{1/3}=(1/8)^{1/3} =1/2

R_{1}: R_{2}=1:2

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

6) N/N_{0}=(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^{- λt}½

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) Y^{4}3 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) =1/4N_{0}

Using N= N_{0} (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=R_{0}e^{-λt}

2700=4750 e_{-5t}

λ=0.113min_{-1}

b) Using T_{1/2}=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.

^{210}Bi_{83}-------------------- ^{210}Po_{84} + ^{0}β_{-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

T_{1/2}(A) = T_{1/2}(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 ΔE_{b}=Δmx931.5 MeV

26) a) ^{226}Ra_{88}--------------→222Rn_{86} +^{4}He2

b) ^{32}P_{15}------------------→^{32}S1_{6}+^{0}e_{-1} +γ

c) ^{32}P_{15}------------------→^{11}B_{5} +^{0}e_{+1} + γ

27) i) ^{6}Li_{3} +^{1}n0 ----------→^{3}H_{1} + ^{4}He_{2} +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)