KCET · Physics · Dual Nature of Matter
The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is ..... (radius of the first orbit of hydrogen atom \(=0.53 Å\) ).
- A \(1.67 \mathrm{~A}\)
- B \(3.33 Å\)
- C \(1.06 Å\)
- D \(0.53 Å\)
Answer & Solution
Correct Answer
(B) \(3.33 Å\)
Step-by-step Solution
Detailed explanation
According to Bohr's quantisation of angular momentum
\[
m v r=\frac{n h}{2 \pi}
\]
or
\(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}}{\mathrm{n}} \quad \text{...(i)}\)
de- Broglie wavelength
\[
\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad \text{...(ii)}
\]
From Eqs. (i) and (ii), we get
Wavelength \(\lambda=\frac{2 \pi \mathrm{r}}{\mathrm{n}}\)
\[
\begin{aligned}
&=\frac{2 \times \pi \times 0.53 Å}{1} \\
&=3.33 Å
\end{aligned}
\]
\[
m v r=\frac{n h}{2 \pi}
\]
or
\(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}}{\mathrm{n}} \quad \text{...(i)}\)
de- Broglie wavelength
\[
\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad \text{...(ii)}
\]
From Eqs. (i) and (ii), we get
Wavelength \(\lambda=\frac{2 \pi \mathrm{r}}{\mathrm{n}}\)
\[
\begin{aligned}
&=\frac{2 \times \pi \times 0.53 Å}{1} \\
&=3.33 Å
\end{aligned}
\]
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