Single Choice

The stopping potential $$(V_0)$$ versus frequency ($$\nu$$) plot of a substance is shown in figure, the threshold wavelength is :

A$$5 \times 10 ^{14}\mathring { A } $$
B$$6000 \mathring { A } $$
Correct Answer
C$$5000 \mathring { A } $$
DCannot be estimated from given data

Solution

From the graph, when $$V_0=0$$, the corresponding frequency is the threshold frequency. i.e $$\nu_0=5\times 10^{14}$$ Hz

so, threshold wavelength $$, \lambda_0=\dfrac{c}{\nu_0}=\dfrac{3\times 10^8}{5\times 10^{14}}=6\times 10^{-7} m=6000 \mathring { A } $$

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