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A nucleus X, initially at rest , undergoes alpha dacay according to the equation , ` _(92)^(A) X rarr _(2)^(228)Y + a ` (a) Find the value of `A` and `Z` in the above process. (b) The alpha particle producted in the above process is found to move in a circular track of radius `0.11 m` in a uniform mmagnatic field of `3` Tesia find the energy (in MeV) released during the process and the binding energy of the patent nucleus X Given that `: m (gamma) = 228.03 u, m(_(0)^(1)) = 1.0029 u. ` `m (_(2)^(4) He) = 4.003 u , m (_(1)^(1) H) = 1.008 u ` |
Answer» Correct Answer - A::B::C::D (a)`_(92)^(A) X rarr _(x)^(228) Y + _(2)^(4)He` `A = 228 + 4 = 232.92 = z = 2 rArr Z = 90` (b) Let v be the velocity with which a particle is emitted Then ` (m nu^(3))/(r ) = q nu B rArr nu = (q r B)/(m) = (2 xx 1.6 xx 10^(-19) xx 0.11 xx 3)/(4.003 xx 10^(-27))` `rArr nu = 1.59 xx 10^(7) ms^(-1)` Appling law of conservation of linear momentum during a - decay we get `m_(1) nu_(1) = m_(0)nu_(0)` ....(i) The total kinetic energy of a particle and `Y` is `E = K.E_(0) = + K.E_(y) = (1)/(2) m_(0) nu_(0)^(2) + (1)/(2) m_(1)nu_(1)^(2)` ` = (1)/(2)m_(0) nu_(0)^(2) + (1)/(2) m nu [(m_(0)nu_(0))/(m_(y))]^(2) = (1)/(2) m_(0)nu_(0)^(2) + m+(0)^(2) + (m_(0)^(2) nu _(0)^(2))/(2 m y)` `= (1)/(2) m_(0) nu_(0)^(2) [1 +(m_(0))/(m nu)]` `= (1)/(2) xx 4.003 xx 1.6 xx 10^(-27) xx (1.59 xx 10^(2)) ^(2) [ 1 + (4.003)/(228.03)]J` `= 8.55 xx 10^(-13) J` `= 5.34 MeV` :. ` Mass equivalent of this energy ` = (5.34)/(931.5) = 0.0051 a m u ` Also `m_(2) = m_(y) + m_(a) + ` mass equivalent of energy (E) `= 228.03 + 4.003 = 0.0057 = 232.0387 a` The number of nucleus `= 92 protons + 140 nutroun ` `:.` Bimdimng energy of nucleus X `= (92 xx 1.006 + 140 xx 1.009) - 232.0387 = 1.971 u` `= = 1.9971 xx 931.5 = 1023 eN` |
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