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 5.1 eV, the stopping potential is estimated to be (the energy of the electron in n text th state .En=(-13.6/n2) e V)

Question

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 5.1 eV, the stopping potential is estimated to be (the energy of the electron in n text th state .En=(-13.6/n2) e V) (A) 5.1 V (B) 12.1 V (C) 17.2 V (D) 7 V

Tardigrade Admin · Accepted Answer

Energy released when electron in the atom jumps from excited state

(n = 3)

to ground state

(n = 1)

is

E = h v = E_{3} - E_{1} = \frac{- 13.6}{3 ^{2}} - (\frac{- 13.6}{1 ^{2}})

= \frac{- 13.6}{9} + 13.6 = 12.1 e V

Therefore, stopping potential

e V_{0} = h v - ϕ_{0}

= 12.1 - 5.1 [∵ work function ϕ_{0} = 5.1]

V_{0} = 7 V

5.1 V

12.1 V

17.2 V

7 V

Energy released when electron in the atom jumps from excited state (n=3) to ground state (n=1) is E=hv=E3​−E1​=32−13.6​−(12−13.6​) =9−13.6​+13.6=12.1eV Therefore, stopping potential eV0​=hv−ϕ0​ =12.1−5.1[∵ work function ϕ0​=5.1] V0​=7V