For the radioactive nuclei that undergo either `alpha or beta` decay, which one of the following cannot occur?

Options

A. Isobar of original nucleus is produced
B. Isotope of original nucleus is produced
C. Nuclei with higher atomic number that of the original nucleus is produced
D. Nuclei with lower atomic number than that of the original nucleus is produced

Answer

C

Explanation

By using, `N=N_(0)((1)/(2))^(1//t_(1//2))`
`implies t=(T_(1//2)log_(e)((N_(0))/(N)))/(log_(e)(2))`
`implies tproplog_(e)(N_(0))/(N) implies (t_(1))/(t_(2))=(log_(e)((N_(0))/(N))_(1))/(log_(e)((N_(0))/(N))_(2))`
Hence, `(6)/(10)=(log_(e)(8//1))/(log_(e)(N_(0)//N))`
`implies log_(e)(N_(0))/(N)=(10)/(6)log_(e)(8)=log_(e)32`
`implies (N_(0))/(N)=32`
So, fraction that decays`=1-(1)/(32)=(31)/(32)`