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  • The threshold frequency of the metal of the cathode in a photoelectric cell is <img alt="1 \times 10^{16} \mathrm{~Hz}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20%5Ctimes%2010%5E%7B1

The threshold frequency of the metal of the cathode in a photoelectric cell is 1 \times 10^{16} \mathrm{~Hz}. When a certain beam of light is incident on the cathode, it is found that a stopping potential 4.144 \mathrm{~V} is required to reduce the current to zero. The frequency of the incident radiation is: (Take \mathrm{h}=6.63 \times 10^{-34} \mathrm{Js} )
 

Option: 1

2.5 \times 10^{15} \mathrm{~Hz}


Option: 2

2 \times 10^{15} \mathrm{~Hz}


Option: 3

4.144 \times 10^{15} \mathrm{~Hz}


Option: 4

3 \times 10^{16} \mathrm{~Hz}


Answers (1)

best_answer

Here,
Threshold frequency, \nu_0=1 \times 10^{15} \mathrm{~Hz}
Stopping potential, \mathrm{V}_{\mathrm{s}}=4.144 \mathrm{~V}
Work function, \phi_0=\mathrm{h} \nu_0

=\frac{6.63 \times 10^{-34} \times 1 \times 10^{15}}{1.6 \times 10^{-19}} \mathrm{eV}=4.144 \mathrm{eV}

According to Einstein's photoelectric equation

\begin{aligned} & \mathrm{h\nu}=\phi_0+\mathrm{eV}_{\mathrm{s}} \\ & =4.144 \mathrm{eV}+4.144 \mathrm{eV}=8.288 \mathrm{eV} \end{aligned}

Energy of incident photon, \mathrm{E}=\mathrm{h\nu}=8.288 \mathrm{eV}

\therefore Frequency of incident photon,

\nu=\frac{E}{h}=\frac{8.288 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{Js}}=2 \times 10^{15} \mathrm{~Hz}
 

Posted by

Rakesh

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