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$βΞ»_{0}=2.3Γ(1.6Γ10_{β19})(6.63Γ10_{β34})Γ3Γ10_{8}β=5.404Γ10_{β7}m$.

$βΞ»_{0}=5404Aβ$

Hence, wavelength $4144Aβ$ and $4972Aβ$ will emit electron from the metal surface.

For each wavelength energy incident on the surface per unit time

$=$ intensity of each $Γ$ area of the surface wavelength

$=33.6Γ10_{β3}βΓ(1Βcm)_{2}=1.2Γ10_{β7}joule$

Therefore, energy incident on the surface for each wavelengths is $2s$

$E=(1.2Γ10_{β7})Γ2=2.4Γ10_{β7}J$

Number of photons $n_{1}$ due to wavelength $4144Aβ$

$n_{1}=(6.63Γ10_{β34})(3Γ10_{8})(2.4Γ10_{β7})(4144Γ10_{β10})β=0.5Γ10_{12}$

Number of photon $n_{2}$ due to the wavelength $4972Aβ$

$n_{2}=(6.63Γ10_{β34})(3Γ10_{8})(2.4Γ10_{β7})(4972Γ10_{β10})β=0.572Γ10_{12}$

Therefore total number of photoelectrons liberated in $2s$,

$N=n_{1}+n_{2}$

$=0.5Γ10_{12}+0.575Γ10_{12}$

$=1.075Γ10_{12}$.

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