Solution
a) changing work function into joules
ϕ=2.7×1.6×10−19\phi=2.7\times 1.6\times10^{-19}ϕ=2.7×1.6×10−19
ϕ=4.32×10−19J\phi=4.32\times10^{-19}Jϕ=4.32×10−19J
b) for threshold frequency
ϕ=hν0\phi=h\nu_0ϕ=hν0
ν0=ϕh=4.32×10−196.6×10−34=65.45×10−15Hz\nu_0=\frac{\phi}{h}=\frac{4.32\times10^{-19}}{6.6\times10^{-34}}=65.45\times10^{-15} Hzν0=hϕ=6.6×10−344.32×10−19=65.45×10−15Hz
C) now maximum wavelength
λ=cν0=3×10865.45×1015=45.83×10−10m\lambda=\frac{c}{\nu_0}=\frac{3\times10^8}{65.45\times10^{15}}=45.83 \times 10^{-10}mλ=ν0c=65.45×10153×108=45.83×10−10m
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