In an experiment of photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.2 \times 10^{- 15} V s$ . Calculate the value of Plank's constant.

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Solution

Formula,

$h v = e V$
$h = e \times \frac{V}{v}$
$\Rightarrow h = 1.6 \times 10^{- 19} \times 4.2 \times 10^{- 15}$
$∴ h = 6.72 \times 10^{- 34} J - s$

Hence, this is the answer.

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In an experiment of photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.2×10−15Vs. Calculate the value of Plank's constant.

Formula,hv=eVh=e×Vv⇒h=1.6×10−19×4.2×10−15∴h=6.72×10−34 J−sHence, this is the answer.

In an experiment of photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.2 \times 10^{- 15} V s$ . Calculate the value of Plank's constant.

Formula,

$h v = e V$
$h = e \times \frac{V}{v}$
$\Rightarrow h = 1.6 \times 10^{- 19} \times 4.2 \times 10^{- 15}$
$∴ h = 6.72 \times 10^{- 34} J - s$

Hence, this is the answer.