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

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

A^{0}

= 3 \times 10^{8}

The work function of a metal is 4.2 eV, its threshold wavelength will be

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h = 6.62 \times 10^{- 34} J s, c = 3 \times 10^{8}

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