The work function of a metallic surface is 5. | vedclass

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc = 12420 \ eV \ \mathring{A}$ (or $12375 \ eV \ \mathring{A}$ as p

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$1.2$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc = 12420 \ eV \ \mathring{A}$ (or $12375 \ eV \ \mathring{A}$ as per the provided constant $h = 4.14 	imes 10^{-15} \ eV \ s$ and $c = 3 	imes 10^8 \ m/s$):$E = \frac{4.14 	imes 10^{-15} 	imes 3 	imes 10^8 	imes 10^{10}}{2000} \ eV = 6.21 \ eV$.Using the photoelectric equation $E = W_0 + K_{max}$,where $K_{max} = eV_0$:$6.21 \ eV = 5.01 \ eV + eV_0$.$eV_0 = 6.21 \ eV - 5.01 \ eV = 1.20 \ eV$.Therefore,the stopping potential $V_0 = 1.2 \ V$.

The work function of a metallic surface is $5.01 \ eV$. The photo-electrons are emitted when light of wavelength $2000 \ \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............. $volts$ $[h = 4.14 \times 10^{-15} \ eV \ s]$

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