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(B) The threshold energy (work function) is given by $\Phi = h 
u_0$. Given $h = 6.6 	imes 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 	imes 10^{-19} \

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$A$ alone

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(B) The threshold energy (work function) is given by $\Phi = h u_0$. Given $h = 6.6 imes 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 imes 10^{-19} \ J$.For metal $A$:$\Phi_A = h u_A = (6.6 imes 10^{-34}) imes (1.8 imes 10^{14}) \ J = 11.88 imes 10^{-20} \ J$.Converting to $eV$: $\Phi_A = \frac{11.88 imes 10^{-20}}{1.6 imes 10^{-19}} \ eV = 0.7425 \ eV$.For metal $B$:$\Phi_B = h u_B = (6.6 imes 10^{-34}) imes (2.2 imes 10^{14}) \ J = 14.52 imes 10^{-20} \ J$.Converting to $eV$: $\Phi_B = \frac{14.52 imes 10^{-20}}{1.6 imes 10^{-19}} \ eV = 0.9075 \ eV$.The energy of incident photons is $E = 0.825 \ eV$.Since $E > \Phi_A$ $(0.825 \ eV > 0.7425 \ eV)$ and $E < \Phi_B$ $(0.825 \ eV < 0.9075 \ eV)$,photoelectrons are emitted only from metal $A$.

$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \ Hz$ and $2.2 \times 10^{14} \ Hz$. Two identical photons of energy $0.825 \ eV$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 10^{-34} \ J \cdot s$)

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