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

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Question

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $h = 4.14 	imes 10^{-15} \ eV \cdot s$ and $c = 3 	imes 10^8 \ m/s$

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$1.2$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $h = 4.14 	imes 10^{-15} \ eV \cdot s$ and $c = 3 	imes 10^8 \ m/s$,the product $hc \approx 12420 \ eV \cdot \mathring{A}$.$E = \frac{12420 \ eV \cdot \mathring{A}}{2000 \ \mathring{A}} = 6.21 \ eV$.According to Einstein's photoelectric equation,$E = W_0 + K_{max}$,where $K_{max} = eV_0$.$V_0 = \frac{E - W_0}{e} = \frac{6.21 \ eV - 5.01 \ eV}{e} = 1.2 \ V$.Thus,the stopping potential is $1.2 \ V$.

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

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