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If the wavelength of the light falling on a surface is increased from 3000 \, \, \AA to 3040\, \, \AA, then what will be the corresponding change in the stopping potential? (Given \, \, that \, \, hc =12.4 \times 10^3 \, \, \mathrm{eV} \, \AA )

Option: 1

5.5 \times 10^{-2} \mathrm{~V}


Option: 2

3.5 \times 10^{-2} \mathrm{~V}


Option: 3

2.75 \times 10^{-2} \mathrm{~V}


Option: 4

None of these


Answers (1)

best_answer

If V_{\circ} represents the stopping potential for a particular wavelength \lambda, then from Einstein's photoelectric equation

\mathrm{eV}_0=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}

(W = Work function of the metal over which light is incident)
As W is a constant for a given metallic surface, hence
differentiating above eq.,
we have:

\operatorname{edV}_0=-\frac{\mathrm{hc} \cdot \mathrm{d} \lambda}{\lambda^2}=-\frac{\left(12.4 \times 10^3 \mathrm{eV} \AA\right)(40 \AA)}{(3000 \AA)^2}

=-0.055 \mathrm{eV}

\therefore \quad \mathrm{dV}_{\circ}=-5.5 \times 10^{-2} \text { volt. }

Posted by

shivangi.shekhar

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