Question 10
In the time period given, source X undergoes five half-lives and source Y undergoes three. The activity is proportional to the number of nuclei present so the activity of source X goes down by a factor of 32 and source Y by a factor of 8.
$\frac{320}{32}=10$
$\frac{480}{8}=60$
$10+60=70Bq$
In the time period given, source X undergoes five half-lives and source Y undergoes three. The activity is proportional to the number of nuclei present so the activity of source X goes down by a factor of 32 and source Y by a factor of 8.
$\frac{320}{32}=10$
$\frac{480}{8}=60$
$10+60=70Bq$