The electron in the hydrogen atom jumps from excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1eV, the stopping potential is estimated to be: (The energy of the electron in nth state is `E_(n)=-13.6//n^(2)eV`)

Options

A. 5.1 V
B. 12.1 V
C. 17.2 V
D. 7 V

Answer

D

Explanation

`DeltaE=13.6((1)/(1^(2))-(1)/(3^(2))=(8)/(9))=12.1 eV`
` E=phi+eV_(s) implies 12.1=5.1+eV_(s)`
`V_(s)=7 V`