Class 12 NUCLIE
NUCLEI
✤ Nucleus was discovered in 1911 by Rutherford.
Size of nucleus is 10⁻¹⁵ m.
# ATOMIC mass is measured by mass spectrometer.
Atomic mass is measured in atomic mass unit (a.m.u)
1 a.m.u = 1.66 x 10⁻²⁷ kg.
1 a.m.u = 931 Mev
✤ NUCLEAR size : Experimental verification shows that Volume of nucleus is proportional to its mass number 'A'. If R is radius of nucleus then
Volume α A ==> 4/3 πR³ α A ==> R³ α A
R α (A)¹/³
R = R₀(A)¹/³ Here R₀= empirical constant
= 1.2 x 10⁻¹⁵m
Ex :- Obtain the ratio of nuclear radii of ₂₆Fe and ₉₂U ?
Sol. R₁ / R2 = (A1/A2)1/3
R₁ = (56/238)1/3 = (0.235)1/3= 0.617.
✤ NUCLEAR density : It is defined as the ratio of mass of nucleus and its volume.
P = mass of nucleus / Volume
P = mA / (4/3) πR³A ==> P = 3m/4 πR³
nuclear density does not depend upon Atomic mass No.
✤ Nuclear binging energy : It is the energy by which nucleons (electrons) are band in the nucleus.
§ Binding energy is used to describe the stability of atom.
§ more binding energy, more stable.
✤ Mass defect : It is the difference in mass of nucleons and mass of nucleus.
Ex - let ZXᴬ is any element then
mass defect = [z x mₚ + (A-z) mₙ] - mN
(ΔΜ)
✤ Binding energy: - It is the product of mass defect and speed of light square.
B.E = ΔM.c² Joules
§ If mass is in a.m.u then Binding energy
B.E = ΔΜ x 931 Mev.
✤ Important feature of B.E curve
1. Except some nuclei like H², H³, He ete, value of B.E curve per nucleon line on a smooth curve.
2. B.E Per nucleon is small for light nuclei
3. more Binding energy Per nucleon, more stable nuclei and vice versa also.
§. on the basis of B-E we can explain nuclear fission as B.E per nuclei is smaller for heavier nuclei than the middle nuclei so heavier nuclei are less stable. When a heavier nucleus splits into lighter nuclei, B.E/nucleon changes from 7.6 Mev to 8.4 Mev. it large amount of energy is librated, it is used in atomic bomb.
❉ Nuclear fusion: The binding energy per nucleon for lighter nuclei is small is. they are less stable. So when two light nuclei combine to form a heavier nuclei, the higher binding energy per nucleon is released it is used in hydrogen bomb.
❉ Radio activity: In our nature there are Some elements which disintegrate continuously these element are called radioactive element and this property is called radioactivity.
§ It is discovered by henry becquerel in 18%.
§ Ex - of radioactive elements → U, Th, Ra, Ru. etc.
Elements having atomic No. greater than 82 shows radioactivity, B-rays, α-rays
✤ Laws of radioactivity:
1. No. of radioactive element (Nuclei) disintegrating per second is directly proponed to the number of atoms present at that time.
i.e dN/dt ∝ N
dN / dt = -λN Here λ is called disintegration constant.
❊ Equation of radioactivity N = No e-λt
N = No. of atoms Present at t' time
No → initial Number of atoms.
2. It is a spontaneous process, we can't predict that which atom is disintegrated first.
✤ HALF life: → It is the time taken by the radioactive element to disintegrate half of its initial atoms.
• It is represented by "T" for half life N = No and t=T
T = 0.693 / λ
Note:
N= No(1/2)n
Here No= initial Number of atoms
N = atoms present after
n = No. of half life. T 'time
n = t / T
Q. A radioactive substance decays to 1/32 of its initial activity in 25 days, calculate half life.
Sol. we have N = No(1/2)n
1/32 No = No (1/2)n
(1/2)5 = (1/2)n
:. n = t/T
T = 25 / 5
Ans --> T= 5 days
✤ Significance of half life: → value of half life given an idea of the Relative stability of a radioactive isotope.
✤ Average life (mean half life):- It is defined as the combine lives of all atoms (Nuclei) to the total Number of atoms present in given sample.
Average life = Total life of all nuclei / No
T = ∫t.dN/N
✤ Relation b/w half life (T) and Average life (z):
we have T = 1 / λ
and T = 0.693 / λ
from this
T= 1/0.693 * T
T= 1.44 T
unit of radioactivity: 1. Rutherford
1 Rd = 106 decay/sec.
2. Curie
1 Ci = 3.7x1010 decay/sec.
3. Becquerel
(S. I unit) 1 Bq = 1 decay/sec.
✤ Properties of a-rays :
1. These are +vely charged helium nucleus (He4)
2. when 1 α is emitted by any element then its atomic No is decreased by 2 unit and mass No. by 4 units.
AXB ---[-α]---> A-2YB-4 + energy(MeV)
3. Penetration Power is minimum
4. ionization power is maximum.
✤ Property of B decay (rays):
1. These are the electrons.
→ antineutrino.
2. When ß- is emitted by any element then atomic No is increased by 1.
AXB ---[-ß]---> A+1YB + energy
3. Penetration Power in ß more than α
4. Ionization Power is less then a.Γ-rays: These are the Packets of energy which are released by Radioactivity element after the emission of ß
particle by which when dement reaches in its excited state.
Note: γ-emission shows no effect on atomic Number and atomic mass No.
✤γ-rays : These are the packets of energy which released by Radioactivity element after the emission of ß particle by which when dement reaches in its external state.
Note :- γ-emission shows no effect on atomic Number and atomic mass No.
✤ Some Imp. Question:
1. What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:4?
Sol: Nuclear density does not depend upon mass No. so ratio of densities are 1:1
2. Why is the mass of nucleus always less than the sum of masses of nucleons?
Sol: When nucleons approach each other in the formation of nucleus, they strongly attract each other so -their P.E. decreases and become –ve. P.E
hold the nucleons in the nucleus. decrement of P.E result in the mass decrement of nucleons.
Q. A --- radio active sample having N nuclei has activity
R --- write down an expression for its half life.
Sol: we know, R = N λ so λ=R/ N and half life T= 0.693/ λ T=0.693 * N /R
Q. A radioactive substance decay to 1/32th of its initial activity in 25 days. calculate half life.
Sol. A.T.Q R = 1/32 Ro
we know R= Ro(1/2)n
1/32= (1/2)n
(1/2)5 = (1/2)n
:. half life = t/n
=25/5=5 year days.
Q. Why do lighter nuclei tends to fuse together?
Sol: when lighter nuclei fuse together, they form heavier nuclei having greater binding energy Per Nucleon and they tend to attain a stable structure.
Q. A radioactive nucleus 'A' undergoes a series
A→A1→A2→A3→A4
the mass and atomic number of A2 are 176 and 71. Determine the mass and atomic No. of A4 and A.
Q. P.E and separation b/w nucleons is given by the graph. by using it give the Answer.
1) force b/w nucleons is strongly repulsive when separation is less than ro.
2) attractive nuclear fore r>ro
Q. The initial concentration of a radioactive substance is No and its half life is 12 hours. what will be its concentration after 36 hours.
Sol. Here half life T= 12 h.
t=36 h
:. No.of half life = t/T = 36/12
we know N=No(1/2)n
= No(1/2)3
=No/8
Q. Which Physical quantity in nuclear reaction is considered equivalent to the value of reaction.
Sol. It is equivalent to the energy absorbed/released in reaction.
Q. Obtain Bohr's quantization Condition for angular momentum of electron orbiting in nth orbit in hydrogen atom on the basis of the wave picture of an electron using de-Broglie hypothesis.
Sol. Let an electron is revolving in orbit of radius 'r’ according to Bohr, a wave associated with electron. Hence circular orbit can be taken
stationary energy state on If it Contains an integral number of de-broglie wave length the we must have 2πr=nλ
by de Broglie λ=h/mv
2πr=nh/mv
Mvr=nh/2π
L=nh/2π
It is the famous Bohr’s quantization condition for 'L'
Q. State Bohr's postulate to define stable orbit in hydrogen atom. How does de brogile's hypothesis Explain the stability of those orbit.
Sol. According to Bohar's postulate for stable orbit, only these circular orbits can be allowed stations state of an electron in which its
angular momentum is an integral multiple of i.e. i = nh / 2π n = 1,2,3,4…..
By de broglie’s λ = h/p = h/mv in stable orbit 2πr = nλ
2πr = nh/mV
Λn = nh/2πmVn (orbital radius of bohr’s model)
Q. Calculate shortest wave length of the bracket series. and state which part of the electro-magnet spectrum does it belong
Sol. for shortest wave length transition of electron ni = oo to nf = 4.
we know 1/λ = R (1/nf2 – 1/nl2)
1/λ = R (1/16 – 1/o)
= R/16
λ = 16/R = 12/1.097X107
λ = 1458.5 m
Q. Write rydberg's formula for wavelength for Sepctral lines of hydrogen atom.
Sol. Rydberg's formula 1 / λ = R(1/n₁² - 1/n₂²)
Q. What is the shortest wave length in Paschen Series.
Sol. In Paschen n₁=3, n₂=∞
1/λ = R (1/9 - 1/∞)
λ= 9/R => λ= 9/1.097x107 = 8204.1 Å
Q. Obtain bohr's radius and ground state energy of a muonic hydrogen atom.
Sol. in Bohr's model. r = n²h²/4π²mkze² i.e r ∝ 1/m
we know in muonic hydrogen, mass = 207 Me
revolve around the proton.
So ru /re = me / mu = 207
ru = 1/207 × re
ru = 1/207 x 0.53 x10⁻¹⁰ m
ru = 2.5 x10⁻¹³ m.
Now, E ∝ m
Eu / Ee = mu / me = 207me / me
Eu = 207 Ee = -20.7x13.6 ev
= -2.8 kev.
Q. For a radioactive substance, show the variation of the total mass disintegrated as a function of time t' graphically.
(ii) If initial mass of radioactive substance is 3.2 mg. It has a half life of 4 h. find the mass of the substance left undecayed after 8h.
Sol. IInd T=4h (given)
mass of substance Left = 1/2 x3.2 = 1.6 mg
mass of substance after next 4 h = 1/2 x1.6= 0.8 mg
Q. There are 4√2 x10⁶ radioactive nuclei in a given radioactive sample. If the half life of sample is 20 second - how many nuclei will decay in
10 second.
Sol. Here No=4√2 x10⁶
F₀ST = 20 sec.
t = 10 Sec.
n = t/T = 10/20 = 1/2
we know N = No (1/2)ⁿ
W = 4 x10⁶x (1/2)½
W = 4x10⁶ Nuclei
✤ REVISION
1. Einstein-energy-mass Equation E=mc²
2. 1 amu = 1/12 x mass of C-12 atom
3. Nuclear radius, R=Ro (A)1/3. Here Ro=1.2x10-13m
4. Nuclear density = Nuclear mass / Nuclear volume = 3m/4πR³
5. 1 amu = 1.66x10-27 kg = 931 MeV
1. ev = 1.6 X10-19 J.
6. Equation of radio activity dN/dt = -λN or N=Noe-λt disintegration constant. initial atom or mass
7. Half life T = loge2 / λ or T = 0.693 / λ
8. N = No(1/2)^n -> No. of half life, initial atom.
9. Mean life τ= 1/λ -> disintegration constant. or τ = 1.44 T -> half life.
10. Decay rate or activity of a substance R = |dN/dt| = λN.
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