Q124 / 125 · BITSAT 2011 · Physics

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

  • Correct answer
    A
    $+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{~m}$
  • B
    $-3.4 \mathrm{eV}, 1.99 \times 10^{-9} \mathrm{~m}$
  • C
    $2.8 \mathrm{eV}, 2.38 \times 10^{-10} \mathrm{~m}$
  • D
    $1.1 \mathrm{eV}, 1.28 \times 10^{-9} \mathrm{~m}$

Solution

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

Hence, the correct answer is option: 'A'.

Relevant Questions