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
B
1.2 × 1020 J
C
6 × 1019 J
D
6 × 1012 J

Solution

## The correct option is C 6 $$\times$$ 10$$^{-19}$$ JAs 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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