$\sqrt p: p $ is a prime number is uncountable

Set of primes is countable and set of irrational is uncountable.

$ \sqrt p$ is an irrational number as p is not perfect square. So given set would be subset of irrationals. Further if i make a function $f(p) = \sqrt p$ would that make a bijective? if so, then countable right? pls clarify