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 $hc = 12420 \, eV \cdot \mathring{A}$ (or $12375 \, eV \cdot \mathrin

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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 $hc = 12420 \, eV \cdot \mathring{A}$ (or $12375 \, eV \cdot \mathring{A}$ as per given constants),we calculate $E = \frac{12375}{2000} \, eV = 6.1875 \, eV$.According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $W_0$ is the work function and $K_{\max}$ is the maximum kinetic energy.$K_{\max} = E - W_0 = 6.1875 \, eV - 5.01 \, eV = 1.1775 \, eV$.The stopping potential $V_s$ is related to the maximum kinetic energy by $K_{\max} = e V_s$.Therefore,$V_s = 1.1775 \, V$,which is approximately $1.2 \, V$.

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 ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

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