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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Physics
Chapter
Dual Nature of Matter

BITSAT 2011 · Physics · Dual Nature of Matter

An electron, in a hydrogen-like atom, is in an excited state. It...

An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of −3.4eV-3.4 \mathrm{eV}. The kinetic energy and the de-Broglie wavelength of the electron are respectively

Options

  1. A+3.4eV,0.66×10−9m+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{ m}
  2. B−3.4eV,1.99×10−9m-3.4 \mathrm{eV}, 1.99 \times 10^{-9} \mathrm{ m}
  3. C2.8eV,2.38×10−10m2.8 \mathrm{eV}, 2.38 \times 10^{-10} \mathrm{ m}
  4. D1.1eV,1.28×10−9m1.1 \mathrm{eV}, 1.28 \times 10^{-9} \mathrm{ m}

Correct answer

A. +3.4eV,0.66×10−9m+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{ m}

Step-by-step solution

En=−3.4eVE_{n}=-3.4 \mathrm{eV} The kinetic energyis equal to the magnitude of total energy in this case. ∴K.E=+3.4eV\therefore \mathrm{K.E}=+3.4 \mathrm{eV} The de Broglie wavelength of electron λ=h2mK=6.64×10−342×9.1×10−31×3.4×1.6×10−19eV=0.66×10−9m\begin{aligned} \lambda=\frac{h}{\sqrt{2 m K}} &=\frac{6.64 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}} \mathrm{eV} \\ &=0.66 \times 10^{-9} \mathrm{ m} \end{aligned}
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