A) isobar of parent
B) isomer of parent
C) isotone of parent
D) isotope of parent
Correct Answer: D
Solution :
Let, the radioactive substance be \[+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{c}^{2}}}{c}\] Radioactive transition is given by \[=\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{{{a}^{2}}}{c}-\frac{{{b}^{2}}}{c}+c \right)=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,(\because c=a+b)\] The atoms of element having same atomic number but different mass numbers are called isotopes. So, \[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\] and \[\therefore \] are isotopes.
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