physics point

  Q1. The radius of Ge nucleus is measured to be twice the radius of 9 4 Be . How many nucleons are there in Ge nucleus?   (a) 64                       (b) 72                      (c) 82                         (d) 86      Ans:   R = R 0 (A) 1/3             R Ge =2* R Be              R 0 (A Ge ) 1/3   = 2*R 0 (9) 1/3             A Ge =8*9=72 .   Q2. The radius of a 64 29 X   nucleus is measured to be   4.8 *10 -13 cm . The radius of a 27 12 Y nucleus can be...

                   Q/1.   The measured mass of deuteron atom ( 2 1 H) , Hydrogen atom ( 1 1 H)   , proton and neutron is 2.01649 u ,1.00782 u , 1.00727 u and 1.00866 u . Find the binding energy of the deuteron nucleus (unit MeV /nucleon).   Ans:   Here A=2, Z=1, N=1   B E = [ZM H + NM N + M ( 1 1 H)] ✖  931.5 MeV           = [1 ✖  1.00782 +1 ✖  1.00866 -2.01649]  ✖ 931.5 MeV            = [0.00238] ✖  931.5 MeV           =2.224 MeV   Q/2. The binding energy of the neon isotope ( 20 10 Ne) is 160.647 MeV. Find its atomic mass?   Ans: Here A=10, Z=10, N=10                M ( A Z X) = [ZM H + NM N ] – B/(931.5 MeV/u)         ...

                                              Bohr Atomic Model In Bohr model, Niles Bohr atom with a positively charged nucleus surrounded by electrons that travel in circular orbits around the nucleus-similar in structure to the solar system, but with attraction provided by electrostatic forces. He suggested that electrons could only have certain classical motions:   1. Electrons in atoms orbit the nucleus.   2. The electrons can only orbit stably, without radiating, in certain orbits (called by Bohr the "stationary orbits"): at a certain discrete set of distances from the nucleus. These orbits are associated with definite energies and are also called energy shells or energy levels. In these orbits, the electron's acceleration does not result in radi...

  Find the Fourier series representing                                                    f (x) = x, 0 < x < 2π and sketch its graph from x = −4π to x = −4π Solution:    Let f (x) =a 0 /2+a 1 cos x + a 2 cos 2x +b 1 sin x +b 2 sin 2x ..... Hence    a 0 =(1/ π)   2 π ∫ 0 x dx =(1/ π)[x^2/2] 0 2 π   = 2π a n = (1/ π) 2 π ∫ 0 f (x) cos nx dx   =(1/ π)[x (sin nx/n) -1 (- cos nx/n 2 ) ] 0 2 π      =(1/ π) [ cos2n π/n 2   - 1/n 2 ]     =(1/n 2 π) [1-1]=0 b n = (1/ π) 2 π ∫ 0 f (x) sin nx dx =(1/ π) 2 π ∫ 0 x sin nx dx =(1/ π) [ x(- cos nx/n 2 )-1 (-sin nx/n 2 )] 0 2 π        =(1/ π) [- 2πcos2nπ/n ]= -2/n ...