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  • A and B are two metals with threshold frequencies <img alt="1.8 \times 10^{14} \mathrm{~Hz}" src="https://entrancecorner.oncodecogs.com/gif.latex?1.8%20%5Ctimes%2010%5E%7B14%7D%20%5Cmathrm%7B%7EHz%

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

Option: 1

B alone

 


Option: 2

 A alone
 


Option: 3

neither A nor B
 


Option: 4

both A and B


Answers (1)

best_answer

\begin{aligned} & \phi_{0_\mathrm{~A}}=\frac{\mathrm{hv}_0}{\mathrm{e}} \mathrm{eV}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(1.8 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}=0.74 \mathrm{eV} \\ & \phi_{0_\mathrm{~B}}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(2.2 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}=0.91 \mathrm{eV} \end{aligned}

Since the incident energy 0.825 \mathrm{eV} is greater than 0.74 \mathrm{eV} and less than 0.91 \mathrm{eV}, so photoelectrons are emitted from metal A only.

Posted by

Gautam harsolia

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