94Be + 11H → ___  +  42He  A. 64LiB. 63BeC. 63He D. 63Li

This is a nuclear reaction and we can solve it by conserving the number of protons and neutrons before and after the reaction.

The notation A ZX represents a nuclide, where:

• Z is the atomic number (number of protons),
• A is the mass number (number of protons + number of neutrons),
• X is the chemical symbol.

For 94Be + 11H → ___ + 42He:

Before the reaction:

• The total number of protons is 9 (from Be) + 1 (from H) = 10.
• The total number of neutrons is 4 (from Be) + 1 (from H) - 9 (total protons) = -4.

After the reaction:

• The total number of protons is 10 - 2 (from He) = 8.
• The total number of neutrons is -4 + 2 (from He) = -2.

So, the missing nuclide must have 8 protons and -2 neutrons. However, the number of neutrons cannot be negative. Therefore, there seems to be a mistake in the given nuclear reaction.