The work function for tungsten and sodium are 4.5 eV and 2.3 eV respectively. If the threshold wavelength $λ$ for sodium is $5460 \circ A$ the value of $λ$ for tungsten is

$5893 \circ A$

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$10683 \circ A$

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$2791 \circ A$

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$528 \circ A$

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Solution

The correct option is C
$2791 \circ A$

Since $W_{0} = \frac{h c}{λ_{0}}; ∴ \frac{(W_{0})_{T}}{(W_{0})_{N a}} = \frac{λ_{N a}}{λ_{T}} o r λ_{T} = \frac{λ_{N a} \times (W_{0})_{N a}}{(W_{0})_{T}} = \frac{5460 \times 2.3}{4.5} = 2791 \circ A$

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The work function for tungsten and sodium are 4.5 eV and 2.3 eV respectively. If the threshold wavelength λ for sodium is 5460∘A the value of λ for tungsten is

The correct option is C 2791∘A Since W0=hcλ0;∴(W0)T(W0)Na=λNaλT or λT=λNa×(W0)Na(W0)T=5460×2.34.5=2791∘A

The work function for tungsten and sodium are 4.5 eV and 2.3 eV respectively. If the threshold wavelength $λ$ for sodium is $5460 \circ A$ the value of $λ$ for tungsten is

The correct option is C
$2791 \circ A$

Since $W_{0} = \frac{h c}{λ_{0}}; ∴ \frac{(W_{0})_{T}}{(W_{0})_{N a}} = \frac{λ_{N a}}{λ_{T}} o r λ_{T} = \frac{λ_{N a} \times (W_{0})_{N a}}{(W_{0})_{T}} = \frac{5460 \times 2.3}{4.5} = 2791 \circ A$