MCQ
What is the de-Broglie wavelength associated with the hydrogen electron in its third orbit
- A$9.96 \times {10^{ - 10}}\,cm$
- ✓$9.96 \times {10^{ - 8}}\,cm$
- C$9.96 \times {10^4}\,cm$
- D$9.96 \times {10^8}\,cm$
✓
Answer
Correct option: B.
$9.96 \times {10^{ - 8}}\,cm$
b
Radius of $3^{rd}$ orbit $r=R \times 3^2=9 R$
Radius of $3^{rd}$ orbit $r=R \times 3^2=9 R$
Since,
$\frac{ nh }{2 \pi}= mvr$
$\frac{ h }{ mv }=\frac{2 \pi r }{ n }$
$\lambda=\frac{ h }{ mv }=\frac{2 \pi \times 9 R }{3}$
$\lambda=6 \pi R$
We know that
$R =0.529\, \mathring A$
Therefore,
$\lambda=6 \pi \times 0.529$
$\lambda=9.96\, \mathring A$
$\lambda=9.96 \times 10^{-8} \,cm$
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