$\begin {array} {1 1} (a)\;6.592 \times 10^{-34} Js & \quad (b)\;6.626 \times 10^{-34} Js \\ (c)\;659.2 \times 10^{-35} Js & \quad (d)\;66.26 \times 10^{-35} Js \end {array}$

Ans : (a) $6.592 \times 10^{-34} Js$

The slope of the cut-off voltage(V) versus frequency(v)

of an incident light is given as:

$V/v = 4.12 \times 10^{-15} Vs$

$V$ is related to frequency by the equation: $hv=eV$

So, $h=e \times V/v = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15} = 6.592 \times 10^{-34} Js$

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