To solve for \( \alpha \), we need to understand how the error probability \( P_e \) is related to the probability of \( N \) exceeding 1. Since \( N \) is a zero-mean Gaussian random variable, the probability \( P(N>1) \) can be computed as:
\[
P(N>1) = 1 - P(N \leq 1) = 1 - \Phi(1),
\]
where \( \Phi(1) \) is the cumulative distribution function (CDF) of the standard normal distribution evaluated at 1.
Using standard normal distribution tables or a calculator, we find:
\[
\Phi(1) \approx 0.8413,
\]
so
\[
P(N>1) \approx 1 - 0.8413 = 0.1587.
\]
Since \( P_e = \alpha P(N>1) \), we can solve for \( \alpha \) by equating it to the known value of \( P_e \), leading to:
\[
P_e = \frac{4}{3} P(N>1),
\]
Thus, the value of \( \alpha \) is \( \frac{4}{3} \).