$\begin {array} {1 1} (a)\;5.6 \times 10^{15}Hz & \quad (b)\;0.56 \times 10^{15}Hz \\ (c)\;0.56 \times 10^{17}Hz & \quad (d)\;5.6 \times 10^{17}Hz \end {array}$

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Ans : (b)

$E = W_o+ E_k$ where, $W_o$ is the work function and $E_k$ is the KE of the

liberated photoelectron

$W_o = 2.3\: eV = 3.68 \times 10^{-19} J$

$E_k = \large\frac{1}{2} mv^2 $

$= \large\frac{1}{2} \times 9 \times 10^{-31} \times (10^4 )^2J = 4.5 \times 10^{-23} J$

So, $E = (3.68 \times 10^{-19} + 4.5 \times 10^{-23} ) J$

$= 3.68045 \times 10^{-19}J$

Frequency, $v= \large\frac{E}{h} = \large\frac{3.68045 \times 10^{-19} }{ (6.63 \times 10^{-34})}$

$= 0.56 \times 10^{15}Hz$

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