CUET Preparation Today
CUET
Physics
Nuclei
a
b
c
d
$ n = \frac{t}{t_1/2} = 4$
$ N = \frac{N_0}{2^n} = \frac{N_0}{16}$
$\text{Amount of substance decayed is } \frac{15N_0}{16}$
$ \text{Percentage decayed } \frac{15N_0/16}{N_0} \times 100 = 93.75$%
