Respuesta :
Answer:
(a) 5.04 eV (B) 248.14 nm (c) [tex]1.21\times 10^{15}Hz[/tex]
Explanation:
We have given Wavelength of the light \lambda = 240 nm
According to plank's rule ,energy of light
[tex]E = h\nu = \frac{hc}{}\lambda[/tex]
[tex]E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV[/tex]
Maximum KE of emitted electron i= 0.17 eV
Part( A) Using Einstien's equation
[tex]E = KE_{max}+\Phi _{0}[/tex], here [tex]\Phi _0[/tex] is work function.
[tex]\Phi _{0}=E - KE_{max}[/tex]= 5.21 eV-0.17 eV = 5.04 eV
Part( B) We have to find cutoff wavelength
[tex]\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}[/tex]
[tex]\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }[/tex]
[tex]\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm[/tex]
Part (C) In this part we have to find the cutoff frequency
[tex]\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz[/tex]
