NEETPhysicsDual Nature of Radiation and MatterMCQ+4 / −1
The light rays having photons of energy 4.2 eV are falling on a metal surface having a work function of 2.2 eV. The stopping potential of the surface is
- A6.4 V
- B2 eV
- C2 V
- D1.1 V
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Correct answer: C
We know, $$K{E_{\max }} = hv - h{v_0}$$
$e{V_0} = hv - h{v_0}$ ($\because$ $$K{E_{\max }} = e{V_0}$$)
$$e{V_0} = 4.2\,eV - 2.2\,eV$$
$\therefore$ ${V_0} = 2\,V$
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