MCQ
If the total energy of an electron in a hydrogen atom in excited state is $-3.4\, eV$, then the de-Broglie wavelength of the electron is
  • A
    $3.3 \times {10^{ - 8}}\,cm$
  • B
    $6.6 \times {10^{ - 10}}\,cm$
  • C
    $3.3 \times {10^{ - 10}}\,cm$
  • $6.64 \times {10^{ - 8}}\,cm$

Answer

Correct option: D.
$6.64 \times {10^{ - 8}}\,cm$
d
$n\lambda  = 2\pi r$

$\lambda  = \frac{{2\pi r}}{n} = \frac{{2 \times 3.14 \times 0.529 \times {{10}^{ - 8}}\,cm \times 4}}{2}$

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