Question 440: Atoms & Nuclei

Question

Acceleration of an electron in the first Bohr's orbit is proportional to $\mathrm{m}=$ mass of electron, $\mathrm{r}=$ radius of the orbit, $\mathrm{h}=$ Planck's constant)

Accepted Answer

The acceleration of an electron in the first Bohr's orbit can be analyzed through the centripetal acceleration formula, which is given by:
$$ a = \frac{v^2}{r} $$
where $ v $ is the tangential velocity of the electron in its orbit, and $ r $ is the radius of the orbit.
In Bohr's model, the angular momentum $ L $ of the electron is quantized and given by:
$$ L = mvr = \frac{nh}{2\pi} $$
where $ m $ is the mass of the electron, $ r $ is the radius of the orbit, $ n $ is the principal quantum number (for the first orbit, $ n = 1 $), and $ h $ is Planck's constant.
Solving for the velocity $ v $ from the angular momentum expression:
$$ v = \frac{h}{2\pi m r} $$
Substitute this back into the centripetal acceleration formula:
$$ a = \frac{v^2}{r} = \frac{\left(\frac{h}{2\pi m r}\right)^2}{r} = \frac{h^2}{4\pi^2 m^2 r^3} $$
Since the constant part $ \frac{1}{4\pi^2} $ is omitted when considering proportionality, the acceleration $ a $ is proportional to:
$$ \frac{h^2}{m^2 r^3} $$
Therefore, the correct option is:
Option B
: $\frac{h^2}{m^2 r^3}$

Answer

$\frac{\mathrm{m}^3 \mathrm{r}^3}{\mathrm{~h}^2}$

Answer

$\frac{h^2}{m^2 r^3}$

Answer

$\frac{\mathrm{h}^2}{\mathrm{mr}^3}$

Answer

$\frac{\mathrm{mr}^3}{\mathrm{~h}^2}$

Atoms & Nuclei

Acceleration of an electron in the first Bohr's orbit is proportional to \mathrm{m}= mass of electron, \mathrm{r}= radius of the orbit, \mathrm{h}= Planck's constant)

Choose the correct answer: