Light of wavelength $1824 \circ A$ , incident on the surface of a metal, produces photo-electrons with maximum energy 5.3 eV. When light of wavelength $1216 \circ A$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metal surface is

3.5 eV

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13.6 eV

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6.8 eV

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1.5 eV

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Solution

The correct option is D
1.5 eV

$E = W_{0} + K_{m a x}$ From the given data E is 6.78 eV $(f o r λ = 1824 \circ A) o r 10.17 e V (f o r λ = 1216 \circ A)$
$∴ W_{0} = E - K_{m a x} = 6.78 - 5.3 = 1.48 e V$
or
$W_{0} = 10.17 - 8.7 = 1.47 e V$

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Light of wavelength $1824 \circ A$ , incident on the surface of a metal, produces photo-electrons with maximum energy 5.3 eV. When light of wavelength $1216 \circ A$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metal surface is

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Light of wavelength 1824 ∘A, incident on the surface of a metal, produces photo-electrons with maximum energy 5.3 eV. When light of wavelength 1216 ∘A is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metal surface is

The correct option is D 1.5 eV E=W0+Kmax From the given data E is 6.78 eV (for λ=1824 ∘A) or 10.17 eV (for λ=1216 ∘A) ∴ W0=E−Kmax=6.78−5.3=1.48eV or W0=10.17−8.7=1.47 eV

Light of wavelength $1824 \circ A$ , incident on the surface of a metal, produces photo-electrons with maximum energy 5.3 eV. When light of wavelength $1216 \circ A$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metal surface is

The correct option is D
1.5 eV

$E = W_{0} + K_{m a x}$ From the given data E is 6.78 eV $(f o r λ = 1824 \circ A) o r 10.17 e V (f o r λ = 1216 \circ A)$
$∴ W_{0} = E - K_{m a x} = 6.78 - 5.3 = 1.48 e V$
or
$W_{0} = 10.17 - 8.7 = 1.47 e V$