Answer:
0.7515875 eV
[tex]4\times 10^{14}\leq f<6.08512\times 10^{14}\ Hz[/tex]
Explanation:
f = Maximum frequency = [tex]7.9\times 10^{14}\ Hz[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
W = Work function = 2.52 eV
Converting to Joules
[tex]W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J[/tex]
Maximum photon energy is given by
[tex]E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J[/tex]
Maximum Kinetic energy is given by
[tex]K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J[/tex]
Converting to eV
[tex]1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV[/tex]
The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV
[tex]W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz[/tex]
The range of frequencies for which no electrons are ejected is
[tex]4\times 10^{14}\leq f<6.08512\times 10^{14}\ Hz[/tex]
