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If a photocell is illuminated with a radiation of 1240 $Å$ , then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength are

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$1 eV, 5200 Å$

$2 eV, 6200 Å$

$3 eV, 7200 Å$

$4 eV, 4200 Å$

answer is B.

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Detailed Solution

$W = h v - e V_{s}$

$\begin{array}{l} h v = energy of incident photon \\ Here h v = \frac{12400}{1240} eV = 10 eV \\ ∴ W = 10 - 8 = 2 eV \end{array}$

$So, λ_{0} = Threshold wavelength$

$λ_{0} = \frac{12400}{2 e V} Å = 6200 Å$

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