MCQ
In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is $[$ Given that Bohr radius, $a_0 = 52.9 \;pm]$
- A$211.6 \;pm$
- ✓$211.6\pi \;pm$
- C$52.9\pi \;pm$
- D$105.8 \;pm$
✓
Answer
Correct option: B.
$211.6\pi \;pm$
b
$\mathrm{n} \lambda=2 \pi \mathrm{r}$
$\mathrm{n} \lambda=2 \pi \mathrm{r}$
$\mathrm{n} \lambda=2 \pi \frac{\mathrm{n}^{2}}{\mathrm{z}} \mathrm{a}_{0}$
$n \lambda=2 \pi \times \frac{n^{2}}{Z} \times 52.9 \mathrm{pm}$
$\lambda=2 \pi \times 52.9 \times 2 \mathrm{pm}$
$=211.6 \pi \mathrm{pm}$
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