$(a)\;5eV,9\times10^{14} Hz\qquad(b)\;2.5eV,9\times10^{14} Hz\qquad(c)\;5eV,18\times10^{14} Hz\qquad(d)\;2eV,18\times10^{14} Hz$

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Answer : (b) $\;2.5eV,9\times10^{14} Hz$

Explanation :

Since $\;V_{s}=2.5V , \quad \; K_{max}=eV_{s}$

So , $\;K_{max}=2.5 eV$

Energy of incident photon

$\varepsilon =\large\frac{12400}{1980}eV=6.26eV$

$W=E-K_{max}=3.76eV$

$h \nu_{th}=W=3.76\times1.6\times10^{-19}J$

$\nu_{th}=\large\frac{3.76\times1.6\times10^{-19}}{6.6\times10^{-34}} \approx 9.1\times10^{14} Hz\;.$

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