We prove that the desired value of $\alpha$ is $\alpha = \frac{2}{51}$.

Assume that some $\alpha < \frac{2}{51}$ satisfies the condition of the problem. Choose nonnegative integer $k \ge 0$ such that $\frac{1}{51 \times 2^k} \le \alpha < \frac{1}{51 \times 2^{k-1}}$. Consider $51 \times 2^k$ pumpkins each having weight $\frac{1}{51 \times 2^k}$ tons. For any distribution of pumpkins in 50 boxes there exists a box with at least two pumpkins and therefore the weight of this box equals at least $\frac{1}{51 \times 2^{k-1}} > \alpha$, a contradiction.

We show that $\alpha = \frac{2}{51}$ satisfies the condition of the problem. Let the number of pumpkins be $m$. Take $m$ empty boxes and put a pumpkin in every one of them. If the two lightest boxes contain in common not more than $\frac{2}{51}$ tons of pumpkins then gather all pumpkins of these two boxes into one of them and remove the empty box. When this operation terminates let the number of boxes be $n$ and they contain $x_1 \le x_2 \le \dots \le x_n$ tons of pumpkins. Thus $x_1 + x_2 > \frac{2}{51}$ and hence $x_2 > \frac{1}{51}$. Therefore
$$1 = x_1 + x_2 + \dots + x_n > \frac{2}{51} + (n-2) \times \frac{1}{51},$$
implying $n < 51$. We have that all the pumpkins are distributed in no more than 50 boxes and there are no more than $\frac{2}{51}$ tons of pumpkins in each box. This completes the proof.