A) \[46\times {{10}^{-19}}J\]
B) \[23\times {{10}^{-19}}J\]
C) \[36.8\times {{10}^{-19}}J\]
D) \[9.2\times {{10}^{-19}}J\]
Correct Answer: C
Solution :
Kinetic energy of both \[\alpha -\]particle and proton = Potential energy of two particles \[=\frac{2e(e)}{4\pi {{\varepsilon }_{0}}r}\] \[=\frac{2\times {{(1.6\times {{10}^{-19}})}^{2}}\times 9\times {{10}^{9}}}{{{10}^{-10}}}\] \[=46.08\times {{10}^{-19}}\] As initial momentum of two particles is zero, their final momentum must also be zero. \[\therefore \] Numerical value of momentum of each particle = p. KE of proton \[=\frac{{{p}^{2}}}{2m}=E\](say) Kinetic energy of \[\alpha -\]particle \[=\frac{{{p}^{2}}}{2(4m)}=\frac{E}{4}\] Total kinetic energy \[=E+\frac{E}{4}=46.08\times {{10}^{-19}}\text{J}\] \[\therefore \] \[E=\frac{4}{5}\times 46\times {{10}^{-19}}\text{J}\] \[=36.8\times {{10}^{-19}}\text{J}\]
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