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A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the terminal voltage V versus (i) R and (ii) the current l. It is found that whenR=4 Omega, the current is 1 A and when R is increased to 9 Omega, the current reduce to 0.5 A. Find the values of the emf E and internal resistance r. |
Answer» Solution :(i)For large impact parameter, almost all alpha particles go nearly undeviated and have small deflection. This shows that the mass of the atom is concentrated in a small volume in the form of nucleus and gives an idea of size of nucleus. (ii)An alpha particle having small impact parameter suffers large scattering and in case of head-on-collision, the alpha particle rebounds back. The radius (size) R of nucleus is RELATED to its mass NUMBER (A) as `R=R_(0)A^(1//3)` If m is the average mass of a nucleon, then mass of nucleus = mA, where A is mass number Volume of nucleus `=(4)/(3)PI R^(3)=(4)/(3)pi (R_(0)A^(1//3))=(4)/(3)pi R_(0)^(3)A` Density of nucleus `rho_(N) = ("mass")/("volume")=(mA)/((4)/(3)pi R_(0)^(3)A)=(m)/((4)/(3)pi R_(0)^(3))=(3M)/(4pi R_(0)^(3))` Clearly nuclear density`rho_(N)` is independent of mass number A. OR  In both the process, mass of nucleus PARTICIPATING in the reaction is greater than mass of the product nuclei. This difference in mass `(Delta m)` is converted into energy and released. Mass defect`= Delta m` Energy released `=Delta m xx 931 MeV` `._(1)^(2)H+._(1)^(3)H rarr ._(3)^(4)He+n` `Delta m = (+2.014102 + 3.016049-1.008665)u` `=0.018883 u` Energy released `= Delta m xx 931.5 MeV = 0.018883xx931 MeV` `= 17.59 MeV` |
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