Question
A platinum surface having work function 5.63 eV is illuminated by a monochromatic source of $1.6 \times 10^{15} Hz$ Hz. What will be the minimum wavelength associated with the ejected electron.
✓
Answer
Given $\emptyset_0=5.63 eV =5.63 \times 1.6 \times 10^{-19} J$
$v=1.6 \times 10^{15} Hz$
$K . E .=h v-\emptyset_0=\frac{h c}{\lambda}$
$\lambda=\frac{h c}{h v-\emptyset_0}$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.63 \times 10^{-34} \times 1.6 \times 10^{15}-5.63 \times 1.6 \times 10^{-19}}$
$=\frac{19.89 \times 10^{-26}}{1.6 \times 10^{-19}(6.63-5.63)}$
$=\frac{19.89 \times 10^{-26}}{1.6 \times 10^{15}}$ $=12.4 \times 10^{-7} m$
$v=1.6 \times 10^{15} Hz$
$K . E .=h v-\emptyset_0=\frac{h c}{\lambda}$
$\lambda=\frac{h c}{h v-\emptyset_0}$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.63 \times 10^{-34} \times 1.6 \times 10^{15}-5.63 \times 1.6 \times 10^{-19}}$
$=\frac{19.89 \times 10^{-26}}{1.6 \times 10^{-19}(6.63-5.63)}$
$=\frac{19.89 \times 10^{-26}}{1.6 \times 10^{15}}$ $=12.4 \times 10^{-7} m$
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