Correct option is D. Total number of angular nodes in all the orbital $$(n-1)^{th}$$ shell is 13.
Number of photons emitted=6
So, $$\dfrac {n(n-1)}{2}=6; n=4$$
excited state is $$3^rd$$ or n=4
photon having highest energy will be $$4 \rightarrow 1$$
so, its energy will be=$$13.6 (\dfrac {1}{1^2}-{1}{4^2})=13.6 \times \dfrac {15}{16}=12.75$$
When it is incident on plate having work function 8eV then KE=12.75 -8=4.75
KE will be equal to this value or may be less if electrons is inner electron. so option (B) is correct.
option (A) is incorrect because all photon have equal velocity which is $$3\times 10^8$$ m/s
(C), (D) also incorrect because number of nodes in $$n^th$$ and $$(n-1)^{th}$$ shall are 6 (Radial node) & 3 (Angular node ) respectively.