Question 395: Atoms & Nuclei

Question

When an electron orbiting in hydrogen atom in its ground state jumps to higher excited state, the de-Broglie wavelength associated with it

Accepted Answer

For an electron in the hydrogen atom, according to BohrÃ¢ÂÂs model,
$$ 2\pi r_n = n\lambda $$
where
$r_n$ = radius of the $n^{	ext{th}}$ orbit
$\lambda$ = de-Broglie wavelength
$n$ = principal quantum number
Also, for hydrogen atom,
$$ r_n \propto n^2 $$
So,
$$ \lambda = \frac{2\pi r_n}{n} $$
Since $r_n \propto n^2$,
$$ \lambda \propto \frac{n^2}{n} = n $$
This means the de-Broglie wavelength increases with increase in orbit number $n$.
When the electron jumps from ground state ($n=1$) to a higher excited state ($n=2,3,\dots$), its de-Broglie wavelength increases.
Therefore, the correct option is:
$$ \boxed{	ext{Option D: will increase}} $$

Answer

will become zero.

Answer

will remain same.

Answer

will decrease.

Answer

will increase.

Atoms & Nuclei

When an electron orbiting in hydrogen atom in its ground state jumps to higher excited state, the de-Broglie wavelength associated with it

Choose the correct answer: