The de-Broglie wavelength of an electron having $80 \mathrm{eV}$ energy is nearly $$ \left(1 \mathrm{eV}=1.6…

The de-Broglie wavelength of an electron having $80 \mathrm{eV}$ energy is nearly $$ \left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right. \text {, } $$ Mass of the electron $=9 \times 10^{-n} \mathrm{~kg}$, Planck's constant $\left.=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
  1. $140 Ã…$
  2. $0.14 Ã…$
  3. $14 Ã…$
  4. $1.4 Ã…$

Solution

Kinetic energy, $\mathrm{KE}=80 \mathrm{eV}$ $\begin{aligned} & =80 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & =128 \times 10^{-19} \mathrm{~J}\end{aligned}$ De-Broglie wavelength $\begin{aligned} \lambda=\frac{h}{\sqrt{2 m(\mathrm{KE})}} & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 128 \times 10^{-19}}} \\ & =\frac{6.6 \times 10^{-34}}{\sqrt{256 \times 9 \times 10^{-50}}} \\ & =\frac{6.6 \times 10^{-34}}{16 \times 3 \times 10^{-25}}=\frac{6.6}{48} \times 10^{-9} \\ & =\frac{66}{48} \times 10^{-10} \\ & =1.4 \times 10^{-10} \mathrm{~m} \\ & =1.4 Ã…\end{aligned}$

Asked in: AP EAMCET 2001

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