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Question

If the threshold wavelength ($$\lambda_0$$) for the ejection of an electron from metal is 330 nm, then work function for the photoelectric emission is:


A
1.2 × 1018 J
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B
1.2 × 1020 J
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C
6 × 1019 J
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D
6 × 1012 J
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Solution

The correct option is C 6 $$\times$$ 10$$^{-19}$$ J
As we know,

$$W=\text{hv}_0 = \displaystyle\frac{hc}{\lambda_0}$$

=$$\displaystyle\frac{6.6 \times 10^{-34} Js \times 3 \times 10^8 m}{330 \times 10^{-9}\ m}$$

= 6 $$\times$$ 10$$^{-19}\ J$$

Hence, the correct option is $$C$$

Chemistry

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