Light of wavelength 4000 Å is incident on a sodium surface for which the threshold wave length of photo – electrons is 5420 Å. The work function of sodium is

4.58 eV

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2.29 eV

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1.14 eV

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0.57 eV

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Solution

The correct option is B
2.29 eV

$W_{0} = \frac{12375}{λ_{0} (Å)} = \frac{12375}{5420} = 2.28 e V$

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Light of wavelength 4000 Å is incident on a sodium surface for which the threshold wave length of photo – electrons is 5420 Å. The work function of sodium is

The correct option is B 2.29 eV W0=12375λ0(Å)=123755420=2.28 eV

The correct option is B
2.29 eV

$W_{0} = \frac{12375}{λ_{0} (Å)} = \frac{12375}{5420} = 2.28 e V$