If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between energies of the two states is radiated in the form of photons. If the work function of the material is `4.2 eV`, then stopping potential is
[Energy of electron in nth orbit `= - (13.6)/(n^(2)) eV`]

Options

A. 2V
B. 4V
C. 6 V
D. 8 V

Answer

C

Explanation

The energy difference between two states is given as,
`DeltaE = E_(2) - E_(1) = (-136)/(2^(2)) = ((-136)/(1^(2)))`
`rArr DeltaE = (13.6)/(1^(2)) - (13.6)/(2^(2))`
`rArr E = 13.6 [(4- 1)/(4)] = 13.6 xx 3/4`
`:. DeltaE = 10.2 eV`
Since, the energy is radiated in form of photons, so we have
enery of phtons `= hv = 10.2 eV`
From Einstein's photoelectric equation, we have
`hv = phi_(0)+ eV_(s)`
Substituting the given values,
`:. 10.2 eV = 4.2 eV + eV_(s)`
`rArr 6 eV = eV_(s)`
`:. V_(s) = 6V`