Question

What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u 1014 /s?

Answer 1

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 ×  $10^{14}$ Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

$E=h\nu_{o}$    ..................1

here E is the energy of electron per atom $E=\frac{x}{N}$  and h is plank constant i.e. $6.626\times10^{-34} Js$  and x is  binding energy

and here N is the Avogadro constant = $6.023\times10^{23}$

so E will $\frac{x}{6.023\times10^{23}}$  

E = $3.19\times10^{-19} J$  

so put value in equation 1 we get

$\frac{x}{6.023\times10^{23}}$ = 2.50 ×  $10^{14}$ × $6.626\times10^{-34} Js$  

solve it we get

x = 99770.99

so  binding energy is 99771 J/mol