An electron in a hydrogen atom jumps from second Bohr orbit to ground state and the energy difference of the two states is radiated in the form of photons. These are then allowed to fall on a metal surface having a work-function equal to text4.2 eV , then the stopping potential is [Energy of electron in nth orbit =-(13 .6/n2) eV ]

Question

An electron in a hydrogen atom jumps from second Bohr orbit to ground state and the energy difference of the two states is radiated in the form of photons. These are then allowed to fall on a metal surface having a work-function equal to text4.2 eV , then the stopping potential is [Energy of electron in nth orbit =-(13 .6/n2) eV ] (A) 2V (B) 4V (C) 6 V (D) 8V

Tardigrade Admin · Accepted Answer

E=(13 .6/1)-(13 .6/22) E=13.6[(4 - 1/4)]=13.6× (3/4) ∴ E=10.2eV (= hÎ½) i.e. hÎ½=10.2eV hÎ½=φ0+eVs ∴ 10.2 eV=4.2 eV+eVs ∴ 6 eV=eVs ∴ Vs=6V

$2 V$

$4 V$

$6 V$

$8 V$

$E = \frac{13.6}{1} - \frac{13.6}{2 ^{2}}$
$E = 13.6 [\frac{4 - 1}{4}] = 13.6 \times \frac{3}{4}$
$∴ E = 10.2 e V (= h ν)$
$i . e . h ν = 10.2 e V$
$h ν = ϕ_{0} + e V_{s}$
$∴ 10.2 e V = 4.2 e V + e V_{s}$
$∴ 6 e V = e V_{s}$
$∴ V_{s} = 6 V$

2V

4V

6V

8V

E=113.6​−2213.6​ E=13.6[44−1​]=13.6×43​ ∴E=10.2eV(=hν) i.e.hν=10.2eV hν=ϕ0​+eVs​ ∴10.2eV=4.2eV+eVs​ ∴6eV=eVs​ ∴Vs​=6V

$2 V$

$4 V$

$6 V$

$8 V$

$E = \frac{13.6}{1} - \frac{13.6}{2 ^{2}}$
$E = 13.6 [\frac{4 - 1}{4}] = 13.6 \times \frac{3}{4}$
$∴ E = 10.2 e V (= h ν)$
$i . e . h ν = 10.2 e V$
$h ν = ϕ_{0} + e V_{s}$
$∴ 10.2 e V = 4.2 e V + e V_{s}$
$∴ 6 e V = e V_{s}$
$∴ V_{s} = 6 V$