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(A) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{W_0}$.Using the approximation $\lambda_0 (	ext{in } \mathring

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Lithium

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(A) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{W_0}$.Using the approximation $\lambda_0 (	ext{in } \mathring{A}) \approx \frac{12400}{W_0 (	ext{in } eV)}$,we calculate the threshold wavelength for both metals.For Lithium $(Li)$: $(\lambda_0)_{Li} = \frac{12400}{2.3} \approx 5391 \ \mathring{A}$.For Copper $(Cu)$: $(\lambda_0)_{Cu} = \frac{12400}{4.0} = 3100 \ \mathring{A}$.The visible light spectrum ranges approximately from $3800 \ \mathring{A}$ to $7500 \ \mathring{A}$.Since $(\lambda_0)_{Li} = 5391 \ \mathring{A}$ falls within the visible region,Lithium can emit photoelectrons when exposed to visible light.Since $(\lambda_0)_{Cu} = 3100 \ \mathring{A}$ falls in the ultraviolet region,Copper cannot emit photoelectrons with visible light.Therefore,Lithium is the correct metal for a photoelectric cell working with visible light.

Work function of lithium and copper are respectively $2.3 \ eV$ and $4.0 \ eV$. Which one of the metal will be useful for the photoelectric cell working with visible light? $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s)$

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