A, Balmer system refers to the photon released when hydrogen atoms jump from high energy level to n=2 energy level. Of the six kinds of photons, n=4→2 and n=3→2 belong to Balmer system, that is, two kinds, so A is correct;

When b, n=4 excited state transitions to the ground state, the photon energy is the largest. According to E=hλC, the wavelength is the shortest, so b is wrong.

C, the photon energy released from the transition from the energy level of n=2 to the ground state is 13.6-3.4= 10.2ev. If a metal plate can have photoelectric effect, the photon energy released from the transition from the energy level of n=3 to the energy level of n=2 is 3.4- 1 = 65438+.

The energy of hydrogen atom with energy level d and n=4 is -0.85ev, so it takes at least 0.85ev to make its ionization energy zero, so d is correct;

E among the six kinds of photons, the photon released from the energy level of n=4 to the energy level of n= 1 has large energy and high frequency, so e is correct;

Therefore: Ade ..