If the work function of a photometal is 6.825 eV. Its threshold wavelength will be $(c = 3 \times 10^{8} m / s)$

$1200 \circ A$

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$1800 \circ A$

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$2400 \circ A$

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$3600 \circ A$

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Solution

The correct option is B
$1800 \circ A$

$λ_{0} = \frac{12375}{6.825} = 1813 \circ A \approx 1800 \circ A$

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Similar questions

If the work function of a photometal is 6.825 eV. Its threshold wavelength will be $(c = 3 \times 10^{8} m / s)$

5040 A^{\circ}

c = 3 \times 10^{8} m / s

6.875

= c = 3 \times 10^{3} m / s

ϕ_{0}

(h = 6.6 \times 10^{- 34} J s, c = 3 \times 10^{8} m s^{- 1}, 1 e V = 1.6 \times 10^{- 19} J)

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