Answer:
binding energy is 99771 J/mol
Exlanation:
given data
threshold frequency = 2.50 × [tex]10^{14}[/tex] Hz
solution
we get here binding energy using threshold frequency of the metal that is express as
[tex]E=h\nu_{o}[/tex] ..................1
here E is the energy of electron per atom [tex]E=\frac{x}{N}[/tex] and h is plank constant i.e. [tex]6.626\times10^{-34} Js[/tex] and x is binding energy
and here N is the Avogadro constant = [tex]6.023\times10^{23}[/tex]
so E will [tex]\frac{x}{6.023\times10^{23}}[/tex]
E = [tex]3.19\times10^{-19} J[/tex]
so put value in equation 1 we get
[tex]\frac{x}{6.023\times10^{23}}[/tex] = 2.50 × [tex]10^{14}[/tex] × [tex]6.626\times10^{-34} Js[/tex]
solve it we get
x = 99770.99
so binding energy is 99771 J/mol
