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# What is the threshold frequency ν0 of cesium? Note that 1 eV (electron volt)=1.60×10−19 J. Express...

What is the threshold frequency ν0 of cesium? Note that 1 eV (electron volt)=1.60×10−19 J. Express your answer numerically in hertz.

Concepts and reason

Calculate the threshold frequency by using the following equation.

${\rm{KE = h}}\nu {\rm{ + h}}{\nu _{\rm{o}}}$

For threshold energy, kinetic energy will be zero.

$\begin{array}{c}\\{\rm{h}}\nu {\rm{ = h}}{\nu _{\rm{o}}}\\\\{\rm{E = h}}{{\rm{\nu }}_{\rm{o}}}\\\end{array}$

Here, h is plank constant, ${\nu _{\rm{o}}}$ is threshold frequency, and E is ionization energy.

Calculate the threshold frequency by using the values of plank constant and the energy.

Fundamentals

Minimum amount of frequency that causes photo electric emission is called as threshold frequency.

Emission of electrons when light is incident on surface of a material is called as photo emission and this effect is called as photo electric effect.

Ionization energy of cesium is ${\rm{375}}{\rm{.7 kJ/mol}}$ .

$\begin{array}{c}\\{\rm{Energy per atom = }}375.7{\rm{ }}\frac{{{\rm{kJ}}}}{{{\rm{mol}}}}{\rm{ }} \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}{\rm{ J}}}}{{{\rm{1 kJ}}}}} \right) \times \left( {\frac{{{\rm{1 mol}}}}{{{\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{ atoms}}}}} \right)\\\\ = {\rm{ }}6.24 \times {10^{ - 19}}{\rm{ J/atom}}\\\end{array}$

Substitute the values in the formula.

$\begin{array}{c}\\6.24 \times {10^{ - 19}}{\rm{ J/atom}} = \left( {{\rm{6}}{\rm{.626 \times 1}}{{\rm{0}}^{ - {\rm{34}}}}{\rm{ }}{{\rm{m}}^{\rm{2}}}{\rm{kg/s}}} \right) \times {\nu _{\rm{o}}}\\\\6.24 \times {10^{ - 19}}{\rm{ J}} \times \frac{{{\rm{kg}}{\rm{.}}{{\rm{m}}^2}/{{\rm{s}}^2}}}{{1{\rm{J}}}} = \left( {{\rm{6}}{\rm{.626 \times 1}}{{\rm{0}}^{ - {\rm{34}}}}{\rm{ }}{{\rm{m}}^{\rm{2}}}{\rm{kg/s}}} \right) \times {\nu _{\rm{o}}}\\\\{\nu _{\rm{o}}} = \frac{{6.24 \times {{10}^{ - 19}}{\rm{kg}}{\rm{.}}{{\rm{m}}^2}/{{\rm{s}}^2}}}{{{\rm{6}}{\rm{.626 \times 1}}{{\rm{0}}^{ - {\rm{34}}}}{\rm{ }}{{\rm{m}}^{\rm{2}}}{\rm{kg/s}}}}\\\\{\nu _{\rm{o}}} = 9.42 \times {10^{14}}{{\rm{s}}^{ - 1}}\\\end{array}$

One cycle per second is equal to hertz. So, the threshold frequency of cesium is $9.42 \times {10^{14}}{\rm{Hz}}$ .

Ans:

Threshold frequency of cesium atom is $9.42 \times {10^{14}}{\rm{Hz}}$ .

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