Hydrogen atoms is a particular excited state 'n', when all returned to ground state, 6 different photons are emitted.Which of the following is/are incorrect .

Options

A. out of 6 different photons only 2 photons have speed equal to that of visible light
B. If highest energy photon emitted from the above sample is incident on the metal plate having work function 8 eV, KE of liberated photo-electron may be equal to or less than 4.75 eV
C. Total number of radial nodes in all the orbitals of `n^(th)` shell is 14
D. Total number of angular nodes in all the orbitals in `(n-1)^(th)` shell is 13

Answer

A,C,D

Explanation

number of photons emitted=6
So, `(n(n-1))/2=6, n=4`
excited state is `3^(rd)` or n=4
photon having highest energy will ge `4to1`
So, its energy will be `=13.6(1/1^2-1/4^2)=13.6xx15/16=12.75`
when it is incident on plate having work function 8 eV then
KE=12.75-8=4.75
KE will be equal to this value or may be less if electron is inner electron.So option (B) is correct.Option (A) is incorrect because all photon have equal velocity which is `3xx10^8m//s`
( C),(D) also incorrect because number of nodes in `n^(th)` & `(n-1)^(th)` shall are 6(Radial node) & 3 (Angular node) respectively.