Question 429: Atoms & Nuclei

Question

The ratio of angular momentum of an electron in $\mathrm{n}^{	ext {th }}$ orbit of hydrogen atom to the velocity of electron in $\mathrm{n}^{	ext {th }}$ orbit is proportional to

Accepted Answer

For hydrogen atom, according to BohrÃ¢ÂÂs model:
Step 1: Angular momentum in the $n^	ext{th}$ orbit
$$ L = m v_n r_n = n \frac{h}{2\pi} $$
Step 2: Velocity of electron in the $n^	ext{th}$ orbit
The velocity in the $n^	ext{th}$ orbit is:
$$ v_n = \frac{2\pi k e^2}{n h} $$
where $k$ is Coulomb's constant and $e$ is the electron charge.
Step 3: Ratio of angular momentum to velocity
We want:
$$ \frac{L}{v_n} $$
From above,
$L = n \dfrac{h}{2\pi}$
$v_n = \dfrac{2\pi k e^2}{n h}$
So,
$ \frac{L}{v_n} = \frac{n \frac{h}{2\pi}}{\frac{2\pi k e^2}{n h}} = n \frac{h}{2\pi} \cdot \frac{n h}{2\pi k e^2} = \frac{n^2 h^2}{4\pi^2 k e^2} $
This ratio is
proportional to $n^2$
.
Final Answer:
Option A: $\boxed{n^2}$

Answer

$\mathrm{n}^2$

Answer

$\frac{1}{\mathrm{n}^2}$

Answer

$\mathrm{n}^3$

Answer

$\frac{1}{\mathrm{n}^3}$

Atoms & Nuclei

The ratio of angular momentum of an electron in \mathrm{n}^{\text {th }} orbit of hydrogen atom to the velocity of electron in \mathrm{n}^{\text {th }} orbit is proportional to

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