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JIPMER 2004
Exam
JIPMER
Year
2004
Subject
Physics
Chapter
Dual Nature of Matter

JIPMER 2004 · Physics · Dual Nature of Matter

The kinetic energy of an electron is 5 eV . Calculate the...

The kinetic energy of an electron is 5 eV . Calculate the de-Broglie wavelength associated with it (h=6.6×10−34Js\left(h=6.6 \times 10^{-34} \mathrm{Js}\right., me=9.1×10−31 kgm_e=9.1 \times 10^{-31} \mathrm{~kg} )

Options

  1. A5.47A˚5.47 Å
  2. B10.9A˚10.9 Å
  3. C2.7A˚2.7 Å
  4. DNone of these

Correct answer

A. 5.47A˚5.47 Å

Step-by-step solution

Wavelength associated with an electron λ=h2mE\lambda=\frac{h}{\sqrt{2 m E}}=6.6×10−342×9.1×10−31×5×1.6×1019=5.47×10−10 m=5.47A˚\begin{aligned} & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{19}}} \\ & =5.47 \times 10^{-10} \mathrm{~m}=5.47 Å\end{aligned}$$
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