In the experiment on photoelectric effect, the graph between E_K(max) and v is found to be a straight line as shown in figure. The threshold frequency and Planck's constant according to this graph are

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$6 \times 10^{18} s^{- 1}, 6 \times 10^{- 34} J - s$

$2.66 \times 10^{18} s^{- 1}, 4 \times 10^{- 34} J - s$

$4 \times 10^{18} s^{- 1}, 3 \times 10^{- 34} J - s$

$3.33 \times 10^{18} s^{- 1}, 6 \times 10^{- 34} J - s$

answer is A.

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Detailed Solution

$\begin{array}{l} K_{max} = h v - W \\ ω is the intercept on y -axis and h is the slope. \\ ∴ h = \frac{2.4 \times 10^{- 15}}{4 \times 10^{18}} = 6 \times 10^{- 34} Js \\ W = 2 \times 10^{- 15} J \\ \Rightarrow h v_{0} = 2 \times 10^{- 15} or v_{0} = 3.33 \times 10^{18} s^{- 1} \end{array}$