Light of wavelength $4000 0 A$ is incident on a sodium surface for which the threshold wavelength of photoelectrons is $5420 0 A$ . The work function of sodium is

0.57 eV

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1.14 eV

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2.29 eV

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4.58 eV

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Solution

The correct option is D 2.29 eV
As we know,

$W = \frac{12375}{λ_{0} (A)} = \frac{12375}{5420} = 2.28 e V$

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Light of wavelength 40000A is incident on a sodium surface for which the threshold wavelength of photoelectrons is 54200A. The work function of sodium is

The correct option is D 2.29 eVAs we know,W=12375λ0(A)=123755420=2.28 eV

Light of wavelength $4000 0 A$ is incident on a sodium surface for which the threshold wavelength of photoelectrons is $5420 0 A$ . The work function of sodium is

The correct option is D 2.29 eV
As we know,

$W = \frac{12375}{λ_{0} (A)} = \frac{12375}{5420} = 2.28 e V$