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Correct option is B)

From Einstein’s equation ,

$eV_{0}=hν–hν_{0}$

$V_{0}=ehν −ehν_{0} $

From graph, threshold frequency of A is less than that of B.

So, as work function = $hν_{0}$

Where $h=Planck’sConstant$

So, work function of A is less than B.

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