"_{46}^{106}Pd(\\alpha, P) _{47}^{109} Ag \\\\\\\\\n _2^4 \\alpha=_2^4 He \\space Nuclei \\space \\& \\space P=proton\\\\\\\\\n_{46}^{106}Pd +_2^4 \\alpha \\to _{47}^{109} Ag +P"
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment