Respuesta :
Answer:
[tex]\lambda=249.2\ nm[/tex]
Explanation:
Given that:
The work function of the rhodium = 480.5 kJ/mol
It means that
1 mole of electrons can be removed by applying of 480.5 kJ of energy.
Also,
1 mole = [tex]6.023\times 10^{23}\ electrons[/tex]
So,
[tex]6.023\times 10^{23}[/tex] electrons can be removed by applying of 480.5 kJ of energy.
1 electron can be removed by applying of [tex]\frac {480.5}{6.023\times 10^{23}}\ kJ[/tex] of energy.
Energy required = [tex]79.78\times 10^{-23}\ kJ[/tex]
Also,
1 kJ = 1000 J
So,
Energy required = [tex]79.78\times 10^{-20}\ J[/tex]
Also, [tex]E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
So,
[tex]79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{79.78\times 10^{-20}}[/tex]
[tex]\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:79.78}[/tex]
[tex]\lambda=\frac{19.878}{10^6\times \:79.78}[/tex]
[tex]\lambda=2.4916\times 10^{-7}\ m[/tex]
Also,
1 m = 10⁻⁹ nm
So,
[tex]\lambda=249.2\ nm[/tex]
