Light of wavelength 1824 $1824 0 A$ , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength $1216 0 A$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

6.8 eV

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1.5 eV

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13.6 eV

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3.5 eV

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Solution

The correct option is A 1.5 eV
As we know,
$E = W_{0} + K_{m a x}$ . From the given data $E$ is $6.75 e V$ (for $λ = 1824 o A$ ) or $10.17 e V$ (for $λ = 1216 o A$ )
$∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V$
OR
$W_{0} = 10.17 - 8.7 = 1.5 e V$

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Light of wavelength 1824 18240A , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength 12160A is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

The correct option is A 1.5 eVAs we know,E=W0+Kmax. From the given data E is 6.75 eV (for λ=1824oA) or 10.17 eV (for λ=1216oA) ∴W0=E−Kmax=6.75−5.3=1.48 eVORW0=10.17−8.7=1.5 eV

Light of wavelength 1824 $1824 0 A$ , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength $1216 0 A$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

The correct option is A 1.5 eV
As we know,
$E = W_{0} + K_{m a x}$ . From the given data $E$ is $6.75 e V$ (for $λ = 1824 o A$ ) or $10.17 e V$ (for $λ = 1216 o A$ )
$∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V$
OR
$W_{0} = 10.17 - 8.7 = 1.5 e V$