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  • Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of <img alt="3 \mathrm{~V}" src="https://entrancecorner.

Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 \mathrm{~V}. photoelectric effect in this metallic surface begins at a frequency 6 \times 10^{14} \mathrm{~s}^{-1} . The frequency of the incident light in \mathrm{s}^{-1} is [Planck's constant =6.4 \times 10^{-34} \mathrm{~J}, charge on the electron =1.6 \times 10^{-19} \mathrm{C} ]
 

Option: 1

7.5 \times 10^{13}


Option: 2

13.5 \times 10^{13}


Option: 3

13.5 \times 10^{14}


Option: 4

7.5 \times 10^{15}


Answers (1)

best_answer

The maximum kinetic energy of emitted photoelectrons
\mathrm{K}_{\max }=\mathrm{eV}_{\mathrm{s}}=\mathrm{e}(3 \mathrm{~V})=3 \mathrm{eV}

Work function, \phi_0=h \nu_0=\left(6.4 \times 10^{-34}\right)\left(6 \times 10^{14}\right) \mathrm{J}
=\frac{\left(6.4 \times 10^{-34}\right)\left(6 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}=2.4 \mathrm{eV}

According to Einstein's photoelectric equation \mathrm{h\nu}=\mathrm{K}_{\max }+\phi_0   where h\nu is incident energy

\mathrm{h\nu}=3 \mathrm{eV}+2.4 \mathrm{eV}=5.4 \mathrm{eV}

\therefore The frequency of the incident light is 

\mathrm{\nu}=\frac{5.4 \mathrm{eV}}{\mathrm{h}}=\frac{5.4 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.4 \times 10^{-34} \mathrm{~J}}=13.5 \times 10^{14} \mathrm{~s}^{-1}

Posted by

Riya

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