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A certain  metal when irradiated by light (r=3.2 x 10¹⁶ Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (r=2.0 x 10¹⁶ Hz)

The v₀ of metal is 

Answer:

Explanation:

(kE)₁ = hv₁- hv₀

 (kE)₂= hv₂-hv₀

 As, (KE)₁ =2 x (KE)₂

$\therefore$    $hv_{1}-hv_{0}=2(hv_{2}-hv_{0})$

or    $hv_{0}=2hv_{2}-hv_{1}$

 or   $v_{0}=2v_{2}-v_{1}$

 $=2\times (2\times 10^{16})-(3.2\times 10^{16})$

 $=0.8\times 10^{16}Hz=8\times 10^{15}Hz$