NEET · Physics · STD 12 - 12. atoms
De-Broglie wavelength of an electron orbiting in the \(n=2\) state of hydrogen atom is close to \((\)Given Bohr radius \(=0.052\text{ nm})\)
- A 0.067 nm
- B 0.67 nm
- C 1.67 nm
- D 2.67 nm
Answer & Solution
Correct Answer
(B) 0.67 nm
Step-by-step Solution
Detailed explanation
Given \(\text n =2, \text Z =1\)
\(2 \pi \text{r} =\text {n} \lambda\)
\(2 \pi \times\left(0.052 \frac{\text{n}^2}{\text{Z}}\right)=\text{n} \lambda\)
On solving \(\lambda=0.67\text{ nm}\)
\(2 \pi \text{r} =\text {n} \lambda\)
\(2 \pi \times\left(0.052 \frac{\text{n}^2}{\text{Z}}\right)=\text{n} \lambda\)
On solving \(\lambda=0.67\text{ nm}\)
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