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Question

# The electric field associated with a light wave is given by $E={E}_{0}\mathrm{sin}\left[\left(1.57×{10}^{7}{\mathrm{m}}^{-1}\right)\left(x-ct\right)\right].$ Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.

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Solution

## Given: $\mathrm{Electric}\mathrm{field},E={E}_{0}\mathrm{sin}\left[\left(1.57×{10}^{7}{\mathrm{m}}^{-1}\right)\left(x-ct\right)\right]$ Work function, $\varphi$ = 1.9 eV On comparing the given equation with the standard equation, $E={E}_{0}\mathrm{sin}\left(kx-wt\right)$, we get: $\omega =1.57×{10}^{7}×c\phantom{\rule{0ex}{0ex}}$ Now, frequency, $v=\frac{1.57×{10}^{7}×3×{10}^{8}}{2\mathrm{\pi }}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}$ From Einstein's photoelectric equation, $e{\mathrm{V}}_{0}=hv-\varphi \phantom{\rule{0ex}{0ex}}$ Here, V0 = stopping potential e = charge on electron h = Planck's constant On substituting the respective values, we get: $e{V}_{0}=6.63×{10}^{-34}×\frac{1.57×3×{10}^{15}}{2\mathrm{\pi }×1.6×{10}^{-19}}-1.9\mathrm{eV}\phantom{\rule{0ex}{0ex}}⇒e{V}_{0}=3.105-1.9=1.205\mathrm{eV}\phantom{\rule{0ex}{0ex}}⇒{V}_{0}=\frac{1.205×1.6×{10}^{-19}}{1.6×{10}^{-19}}=1.205\mathrm{V}$ Thus, the value of the stopping potential is 1.205 V.

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