$\begin {array} {1 1} (a)\;2.54\: eV & \quad (b)\;2.16\: eV \\ (c)\;21.6\: eV & \quad (d)\;2.26\: eV \end {array}$

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Ans : (b)

$\lambda = 488\: nm =488 \times 10^{-9} m$

$V_{\circ} = 0.38 V = 0.38/ (1.6 \times 10^{-19} ) eV$

So, $eV_{\circ} = hc/ \lambda – \phi_{\circ}$

$ \Rightarrow \phi_{\circ} = hc/ \lambda - eV_{\circ}$

$= (6.6 \times 10^{-34} \times 3 \times 10^8 )/ (1.6 \times 10^{-19} \times 488 \times 10^{-9} ) –

(1.6 \times 10^{-10} \times 0.38)/(1.6 \times 10^{-19} )$

$= 2.54 – 0.38 = 2.16 \: eV$

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