Solution
a) changing work function into joules
"\\phi=2.7\\times 1.6\\times10^{-19}"
"\\phi=4.32\\times10^{-19}J"
b) for threshold frequency
"\\phi=h\\nu_0"
"\\nu_0=\\frac{\\phi}{h}=\\frac{4.32\\times10^{-19}}{6.6\\times10^{-34}}=65.45\\times10^{-15} Hz"
C) now maximum wavelength
"\\lambda=\\frac{c}{\\nu_0}=\\frac{3\\times10^8}{65.45\\times10^{15}}=45.83 \\times 10^{-10}m"
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