The work function of sodium is 2.3 eV. The threshold wavelength of sodium will be

2900 \circ A

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2500 \circ A

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5380 \circ A

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2000 \circ A

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Solution

The correct option is D $5380 \circ A$
$ϕ_{0} (e V) = \frac{12375}{λ_{0} (\circ A)}$
$\Rightarrow λ_{0} (\circ A) = \frac{12375}{ϕ_{0} (e V)} = \frac{12375}{2.3} = 5380 \circ A$

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The work function of sodium is 2.3 eV. The threshold wavelength of sodium will be

The correct option is D 5380∘Aϕ0(eV)=12375λ0(∘A)⇒λ0(∘A)=12375ϕ0(eV)=123752.3=5380∘A

The correct option is D $5380 \circ A$
$ϕ_{0} (e V) = \frac{12375}{λ_{0} (\circ A)}$
$\Rightarrow λ_{0} (\circ A) = \frac{12375}{ϕ_{0} (e V)} = \frac{12375}{2.3} = 5380 \circ A$