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The mass of an electron is \(9.1 \times 10^{-31} \mathrm{~kg}\). If its K.E. is \(3.0 \times 10^{-25} \mathrm{~J}\), calculate its wavelength.
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\(K . E .=\frac{1}{2} m v^2\)
\(\therefore v=\sqrt{\frac{2 \mathrm{~K} . \mathrm{E}}{\mathrm{m}}}=\sqrt{\frac{2 \times 3.0 \times 10^{-25} \mathrm{~J}}{9.1 \times 10^{-31} \mathrm{~kg}}}=812 \mathrm{~ms}^{-1}\)
\(\left(1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{m^{2 } \mathrm { s } ^ { - 2 } )}\right.\)
By de Broglie equation,
\(\begin{aligned}
\lambda &=\frac{h}{\mathrm{mv}}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(812 \mathrm{~ms}^{-1}\right)} \\
&=8.967 \times 10^{-7} \mathrm{~m}=8967 Å
\end{aligned}\)
\(\therefore v=\sqrt{\frac{2 \mathrm{~K} . \mathrm{E}}{\mathrm{m}}}=\sqrt{\frac{2 \times 3.0 \times 10^{-25} \mathrm{~J}}{9.1 \times 10^{-31} \mathrm{~kg}}}=812 \mathrm{~ms}^{-1}\)
\(\left(1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{m^{2 } \mathrm { s } ^ { - 2 } )}\right.\)
By de Broglie equation,
\(\begin{aligned}
\lambda &=\frac{h}{\mathrm{mv}}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(812 \mathrm{~ms}^{-1}\right)} \\
&=8.967 \times 10^{-7} \mathrm{~m}=8967 Å
\end{aligned}\)
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