High speed photodiode for laser pulse width measurement
3 to 10 &micro;J / 5 ns = average power of 600-2000 W.

Spot size 1 cm vs. detector size of 0.049 mm2 means that the detector is intercepting about 1/1600 of this power, or about 0.375 to 1.25 W.

This gives a current of 0.375 to 1.25 W * 0.48A/W = 0.18 to 0.6 A

This will charge 0.94 pF at a rate of 0.18 to 0.6 A / 0.94 pF = 0.19 to 0.64 V/ps. Of course, the overall capacitance of your test setup will be at least an order of magnitude higher.