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A metal is irradiated with light of wavelength $660 \mathrm{~nm}$. Given that the work that the work function of the metal is $1.0 \mathrm{eV}$, the de Broglie wavelength of the ejected electron is close to -
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The correct answer is:
$1.3 \times 10^{-9} \mathrm{~m}$
$\begin{array}{l}
\mathrm{E}=\phi+\mathrm{K} . \mathrm{E} \\
\quad \mathrm{E}=\frac{\mathrm{hC}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{660 \times 10^{-9}} \\
=3 \times 10^{-19} \mathrm{~J} \\
\phi=1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J} \\
\text { K.E. }=3 \times 10^{-19}-1.6 \times 10^{-19}=1.4 \times 10^{-19} \mathrm{~J}
\end{array}$
for wave length of emitted electron
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.4 \times 10^{-19}}}=\frac{6.6 \times 10^{-34}}{5 \times 10^{-25}}=1.32 \times 10^{-9} \text { meter }$
\mathrm{E}=\phi+\mathrm{K} . \mathrm{E} \\
\quad \mathrm{E}=\frac{\mathrm{hC}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{660 \times 10^{-9}} \\
=3 \times 10^{-19} \mathrm{~J} \\
\phi=1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J} \\
\text { K.E. }=3 \times 10^{-19}-1.6 \times 10^{-19}=1.4 \times 10^{-19} \mathrm{~J}
\end{array}$
for wave length of emitted electron
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.4 \times 10^{-19}}}=\frac{6.6 \times 10^{-34}}{5 \times 10^{-25}}=1.32 \times 10^{-9} \text { meter }$
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