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If the electron in hydrogen atom jumps...

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`]

A

2eV

B

4eV

C

6eV

D

8eV

Text Solution

Verified by Experts

The correct Answer is:
C
`E=(13.6)/(1)-(13.6)/(2^(2))`
`E=13.6 [(4-1)/(4)]=13.6 xx 3/4`
`thererfore E=10.2eV (=hv)`
i.e, `hv=10.2eV`
`hv=phi_(0)+eV_(S)`
`therefore 6eV=eV_(s)`
`therefore V_(s)=6V`
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