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. 12.1V
C. 17.2V
D. 7V

Answer

D

Explanation

Energy released when electron in the atom jumps from excited state (n = 3) to ground state (n = 1) is
`E = hupsilon = E_3 -E_1 =(-13.6)/3^2-((-13.6)/1^2)`
` = (-13.6)/9 +13.6 = 12.1 eV`
Therefore, stopping potential
`eV_s = hupsilon_0 = 12.1 - 5.1 " "[ :. ` Work function `phi_0 = 5.1` ]
`V_s = 7 V`