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$⇒λ_{0}=2.3×(1.6×10_{−19})(6.63×10_{−34})×3×10_{8} =5.404×10_{−7}m$.

$⇒λ_{0}=5404A∘$

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

For each wavelength energy incident on the surface per unit time

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

$=33.6×10_{−3} ×(1cm)_{2}=1.2×10_{−7}joule$

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

$E=(1.2×10_{−7})×2=2.4×10_{−7}J$

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

$n_{1}=(6.63×10_{−34})(3×10_{8})(2.4×10_{−7})(4144×10_{−10}) =0.5×10_{12}$

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

$n_{2}=(6.63×10_{−34})(3×10_{8})(2.4×10_{−7})(4972×10_{−10}) =0.572×10_{12}$

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

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

$=0.5×10_{12}+0.575×10_{12}$

$=1.075×10_{12}$.

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