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* The binding energy per nucleon is small for smallnuclei.* For 2 < A < 20, there are well defined maximawhich indicate that these nuclei are more stable.* For 30 < A < 120 the average B.E./A is 8.5 MeV /nucleon with a peak value of 8.8 MeV for Iron.* For A > 120, there is a gradual decreases inB.E./nucleon.* More the B.E./A, more is the stability.Radioactivity :β particles are electrons emitted from the nucleus.(n → p + β)(a) N = N 0 e –λt(b)−dNdt⎛ 1 ⎞(c) N = N 0 ⎜ ⎟⎠⎝ 2⎛ 1 ⎞⇒ A = A 0 ⎜ ⎟⎠⎝ 2(d) T 1/2 =(e) τ = λ1dN= λN where = activity leveldtnn0.693λ(f) τ = 1.4 T 1/2(g) t =⎛ 1 ⎞ T 1/ 2= N 0 ⎜ ⎟⎠⎝ 22.303 Nlog 0 10 =λ Ntwhere A = activity level2.303 Alog 0 10λ A2.303 m= log 0λ m(h) If a radioactive element decays by simultaneous−dNemission of two particle then = λ 1 N + λ 2 NdtThe following parameters remain conserved during anuclear reaction(a) linear momentum(b) Angular momentum(c) Number of nucleons(d) Charge(e) The energy released in a nuclear reactionX + P → Y + Z + QQ = [m x + m p ) – (m y + m z )]c 2 = ∆m × c 2Q = ∆m × 931 MeV(f) In a nuclear fusion reaction small nuclei fuse togive big nuclei whereas in a nuclear fusion reaction abig nuclei breaks down.Thermal neutrons produce fission in fissile nuclei.Fast moving neutrons, when collide with atoms ofcomparable masses, transfer their kinetic energy tocolliding particle and slow down.According to Doppler's effect of lightPower, P = tE =η =out putIn putnhν =tnhcλt∆ λλSolved Examples=cv1. The energy of an excited hydrogen atom is –3.4 eV.Calculate the angular momentum of the electronaccording to Bohr theory.Sol. The energy of the electron in the nth orbit is2nE n = –13.6Here, – = –3.413.6eV2nor n = 2nh 2×6.63×10Angular momentum = =2π 2×3. 14= 2.11 × 10 –34 Js.−342. The wavelength of the first member of the Balmerseries in the hydrogen spectrum is 6563 Å. Calculatethe wavelength of the first member of the Lymanseries.Sol. For the first member of the Balmer series1 ⎡ 1 1 ⎤ 5R= Rλ⎢ − ⎥⎣ 2 23 2=⎦ 36For the first member of the Lyman series1λ ´= R ⎡ 1 1 ⎤⎢ − ⎥⎣122 2 ⎦=3R4Dividing Eq. (1) by Eq. (2)λ´ 5×4 5= =λ 36×3 27or λ´ = 275 λ = 275 × 6563 = 1215 Å...(1)...(2)XtraEdge for IIT-JEE 28 APRIL 2010