KCET · Physics · Atomic Physics
The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is
- A \(0.3 Å\)
- B \(3.3 Å\)
- C \(6.26 Å\)
- D \(10 Å\)
Answer & Solution
Correct Answer
(B) \(3.3 Å\)
Step-by-step Solution
Detailed explanation
The angular momentum of electron in an orbit of radius \(r\) is,
\(m v r=\frac{n h}{2 \pi}\)
So, momentum in ground state,
\(p=m v=\frac{h}{2 \pi r}...(i)\)
The de-Broglie wavelength associated with an electron is
\(\lambda=\frac{h}{m v}=\frac{h}{p}=2 \pi r\) [from Eq. (i)]
As we know, in ground state of hydrogen atom, \(r=0.53 Å\)
\(\Rightarrow \quad \lambda=2 \pi \times 0.53 Å=3.3 Å\)
\(m v r=\frac{n h}{2 \pi}\)
So, momentum in ground state,
\(p=m v=\frac{h}{2 \pi r}...(i)\)
The de-Broglie wavelength associated with an electron is
\(\lambda=\frac{h}{m v}=\frac{h}{p}=2 \pi r\) [from Eq. (i)]
As we know, in ground state of hydrogen atom, \(r=0.53 Å\)
\(\Rightarrow \quad \lambda=2 \pi \times 0.53 Å=3.3 Å\)
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