question_answer

The nucleus\[F{{e}^{57}}\]emits a ray of energy 14.4 keV. If the mass of the nucleus is 56.935 u, calculate the recoil energy of nucleus.

Answer:

                Here, \[{{E}_{1}}=14.4keV=14.4\times 1.6\times {{10}^{-16}}J\], \[{{M}_{2}}=56.935\], \[u=56.935\times 1.66\times {{10}^{-27}}kg\] According to the principle of conservation  of linear momentum \[\frac{{{E}_{1}}}{c}+\sqrt{2{{M}_{2}}{{E}_{2}}}=0\]           \[\therefore \]   \[-\frac{{{E}_{1}}}{c}=\sqrt{2{{M}_{2}}{{E}_{2}}}\]     or             \[\frac{{{E}_{1}}^{2}}{{{c}^{2}}}=2{{M}_{2}}{{E}_{2}}\] \[\therefore \]   \[{{E}_{2}}=\frac{E_{1}^{2}}{2{{M}_{2}}{{c}^{2}}}=\frac{{{\left( 14.4\times 1.6\times {{10}^{-16}} \right)}^{2}}}{2\times 56.935\times 1.66\times {{10}^{-27}}{{\left( 3\times {{10}^{8}} \right)}^{2}}}J\] \[{{E}_{2}}=3.1185\times {{10}^{-22}}J=\frac{3.1185\times {{10}^{-22}}}{1.6\times {{10}^{-16}}}keV=1.95\times {{10}^{-6}}keV\]