A) 0.52 keV
B) 0.75 eV
C) 0.52 eV
D) 0.75 keV
Correct Answer: A
Solution :
Binding energy, \[{{T}_{n}}=2{{T}_{i}}\] where \[{{T}_{n}}={{T}_{i}}-T\] Here, \[{{T}_{n}}={{T}_{i}}+T\] \[\beta \] \[\Omega \] For cloud chamber, the range-energy relation is \[\Omega \] or \[\Omega \] \[\Omega \] \[\Omega \] and \[\phi \] Energy of the incident photon \[\phi ={{\phi }_{0}}+4t\] \[\phi \] \[{{\phi }_{0}}\] From Einsteins photoelectric equation \[\frac{T}{4}\] \[\frac{T}{8}\] Binding energy, \[\frac{T}{12}\] or \[\frac{T}{2}\] \[=1.14\,keV\] and \[\lambda \] \[{{\left( \frac{1}{2} \right)}^{2}}\]
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