Question

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is:
1. 3.5 Å
2. 6.6 Å
3. 12.9 Å
4. 9.7 Å

Answer 1

Correct Answer - Option 4 : 9.7 Å

Concept:

De Broglie wavelength λ, is defined as the Planck’s constant is divided by the particle’s momentum.

Formula: \(\lambda = \frac{h}{p}\)

It is used to calculate the wavelength and momentum.

From question, de Broglie's assumption about angular momentum is:

\( \Rightarrow {\rm{mvr}} = \frac{{{\rm{nh}}}}{{2{\rm{\pi }}}}\)

\( \Rightarrow \frac{{2{\rm{\pi r}}}}{{\rm{n}}} = \frac{{\rm{h}}}{{{\rm{mv}}}}\)

We know that, the de Broglie’s wavelength is given by the formula:

\({\rm{\lambda }} = \frac{{\rm{h}}}{{{\rm{mv}}}}\) 

Now, the equation becomes,

\( \Rightarrow {\rm{\lambda }} = \frac{{2{\rm{\pi r}}}}{{\rm{n}}}\) 

Where,

n = Excited state = 3 (given)

r = Radius of the atom = 4.65 Å = 4.65 × 10^-10 m

Calculation:

On substituting values,

\( \Rightarrow {\rm{\lambda }} = \frac{{2{\rm{\pi }}\left( {4.65 \times {{10}^{ - 10}}} \right)}}{3}\) 

⇒ λ = 9.73 × 10^-10 m

∴ λ = 9.7 Å