In a photoelectric effect experiment, stopping potential changes by 30 volt if we change frequency of the radiation. Then the magnitude of change in the frequency is: $(h = 6 \times 10^{- 34} J - s)$

4 \times 10^{15} s^{- 1}

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8 \times 10^{15} s^{- 1}

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10^{16} s^{- 1}

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18 \times 10^{15} s^{- 1}

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Solution

The correct option is D $8 \times 10^{15} s^{- 1}$
$e V_{0} = h_{υ} - h υ_{0}$
$V_{0} = (\frac{h}{e}) υ - (\frac{h}{e}) υ_{0}$
Slope, $m = \frac{h}{e} = \frac{Δ V_{0}}{Δ_{υ}}$
$Δ_{υ} = \frac{e}{h} \times Δ V_{0}$
$= \frac{1.6 \times 10^{- 19}}{6 \times 10^{- 34}} \times 30 = 8 \times 10^{15} s^{- 1}$

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In a photoelectric effect experiment, stopping potential changes by 30 volt if we change frequency of the radiation. Then the magnitude of change in the frequency is: (h=6×10−34J−s)

The correct option is D 8×1015s−1eV0=hυ−hυ0V0=(he)υ−(he)υ0Slope, m=he=ΔV0ΔυΔυ=eh×ΔV0 =1.6×10−196×10−34×30=8×1015s−1

In a photoelectric effect experiment, stopping potential changes by 30 volt if we change frequency of the radiation. Then the magnitude of change in the frequency is: $(h = 6 \times 10^{- 34} J - s)$