Length of prime number used in Pedersen Commitment I am writing a program using a Pedersen commitment scheme and all I'm missing is an appropriate length for my prime $p$. I have heard that a length of $2^{80}$ is ok, is that correct?
If you mean a prime in the neighborhood of $2^{80}$, well, that is incorrect.  A prime that small will allow someone to commit to a value, and then reveal another one.

A Pederson commitment is a value $g^x h^r \bmod p$, where $g$, $h$ and $p$ are public values, $x$ is the value being committed to, and $r$ is a random value.  To reveal the commitment, you publish $x$ and $r$.

To reveal a second value $x'$, one way to do this is compute a value $r'$ with $g^x h^r \equiv g^{x'} h^{r'}$, this is equivalent to $h^{r'} \equiv c$, where the attacker can compute $c \equiv g^{x - x'}h^{r}$.  If the attacker can solve his discrete logarithm to the base $h$, we can recover $r'$ and reveal his second value $x'$.

Now, for primes of 80 bits in length, this discrete logarithm is easily solved.  To make it impractical, you need a prime at least 1024 bits in length, preferably at least 2048 bits.

On the other hand, if you mean a prime that's $2^{80}$ bits in length, that is, around $2^{2^{80}}$, well, you'll find that is an impractically large prime to work with; computing $g^xh^r$ would be infeasible.