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When a surface is irradiated with light of wavelength $$4950 \mathring{A}$$, a photocurrent appears which vanishes if a retarding potential greater than $$0.6 V$$ is applied across the phototude. When a different source of light is used, it is found that the critical retarding potential is changed to $$1.1 V$$. Find the work function of the emitting surface and the wavelength of the second source.

Solution
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We know that
$$hf_1 = \phi + \dfrac{1}{2} mv^2_1 $$ and $$\dfrac{1}{2} mv^2_1 = eV_1$$
$$\phi = hf_1 - eV_1 = \dfrac{hc}{\lambda_1} - eV_1 = \dfrac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{4950 \times 10^{-10}} - (1.6 \times 10^{-19}) (0.6)$$
$$= 3.04 \times 10^{-19} V ; hf_2 = \phi + eV_2$$
$$\dfrac{hc}{\lambda_2} =3.04 \times 10^{-19} + (1.6 \times 10^{-19} ) \times 1.1 $$
$$= 4.8 \times 10^{-19}$$
$$\Rightarrow \lambda_2 = \dfrac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{4.8 \times 10^{-19}} = 4125 \mathring{A}$$.

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