In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{- 15} V s$ . Calculate the value of Plancks constant.

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Solution

The ratio between the cut-off voltage and the frequency of light is given as:
$\frac{C u t o f f v o l t a g e}{f r e q u e n c y o f l i g h t}$ = $\frac{V}{ν} = 4.12 \times 10^{- 15} V s$

Now, the energy of the photoeletron is given as:
$E = h ν = e V$

Therefore, h=e\dfrac{V}{\nu}
$\Rightarrow 1.6 \times 10^{- 19} \times 4.12 \times 10^{- 15}$

So, $h = 6.592 \times 10^{- 34} J s^{- 1}$

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In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×10−15Vs. Calculate the value of Plancks constant.

The ratio between the cut-off voltage and the frequency of light is given as:Cut off voltagefrequency of light=Vν=4.12×10−15VsNow, the energy of the photoeletron is given as:E=hν=eVTherefore, h=e\dfrac{V}{\nu}⇒1.6×10−19×4.12×10−15So, h=6.592×10−34Js−1

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{- 15} V s$ . Calculate the value of Plancks constant.