Answer:
Planck’s constant is [tex]6.9*10^{-34} Js[/tex]
work function of metal is [tex]5.846*10^{-19}[/tex]
Explanation:
The maximum kinetic energy from electron from photoelectric effect is given by
[tex]\K_{max}=eV_o[/tex] where [tex]V_o[/tex] is applied voltage and e is charge on electron and substituting charge on electron by [tex]1.6*10^{-19}[/tex] and 1.00 V for [tex]V_o[/tex]
[tex]K_{max}=1.6*10^{-19}*1=1.6*10^{-19} J[/tex]
Considering that the wavelength, \lambda of light used is given as [tex] 278*10^{-9} m[/tex]
Energy of light is given by
[tex]E=\frac {hc}{ \lambda}[/tex] where [tex]\lambda[/tex] is the wavelength, h is Planck’s constant and c is speed of light. Taking c for [tex]3*10^{8} m/s[/tex]
Substituting values of wavelength and speed of light we obtain
[tex]E=\frac {h*3*10^{8}}{278*10^{-9}}=h1.07914*10^{15} J[/tex]
If work function is [tex] \phi[/tex] then
[tex]E=\phi + k_{max}[/tex] hence
[tex]h1.07914*10^{15} J=\phi +1.6*10^{-19} J[/tex] ------- Equation 1
Energy corresponding to wavelength of 207 nm is
[tex]E=h\frac {3*10^{8} m/s}{207*10^{-9}}=h(1.45*10^{15}}) J[/tex]
Maximum kinetic energy of electrons when [tex]V_o[/tex] is 2.6 V becomes
[tex]K_{max}=(1.6*10^{-19})*2.6V=4.16*10^{-19} J[/tex]
From [tex]E=\phi + k_{max}[/tex] and substituting [tex] h(1.45*10^{15}}) J[/tex] for E and [tex]4.16*10^{-19} J[/tex] for [tex]K_{max}[/tex] we have
[tex]h(1.45*10^{15}}) J=\phi +4.16*10^{-19} J[/tex] ------ Equation 2
Equation 2 minus equation 1 gives
[tex]h3.71*10^{14}=2.56*10^{-19}[/tex]
[tex]h=\frac {2.56*10^{-19}}{3.71*10^{14}}\approx 6.9*10^{-34} Js[/tex]
Therefore, Planck’s constant is [tex]6.9*10^{-34} Js[/tex]
Recalling equation 1 and substituting back the value of h as obtained
[tex]h1.07914*10^{15} J=\phi +1.6*10^{-19} J[/tex]
[tex] (6.9*10^{-34})*(1.07914*10^{15})=\phi +1.6*10^{-19} J[/tex]
[tex]\ph=(6.9*10^{-34})*(1.07914*10^{15})-(1.6*10^{-19})=5.846*10^{-19}[/tex]
Therefore, work function of metal is [tex]5.846*10^{-19}[/tex]
