Question

Derive the velocity of an electron in a hydrogen atom using Bohr's model. The electron has a mass of m, charge of (-e) and an orbit radius of r. Make sure you explain all the teps including quantization of angular momentum to get full credits

Answer 1

Answer:

$v=\frac{Ze^2}{2 \epsilon_0\times n\times h}$

$v=\frac {2.185\times 10^6}{n}\ m/s$

Explanation:

According to Bohr's Theory,

The force of attraction acting between the electron and the nucleus is equal to the centrifugal force acting on the electron.

Thus,

$\frac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\frac{v^2}{r}$ ......1

Where, Z is the atomic number or the number of protons

r is the atomic radius

v is the velocity of the electron

$m_e$ is the mass of the electron

Also,

Accoriding to Bohr, the angular momentum is quantized. He states that the angular momemtum is equal to the integral multiple of $\frac {h}{2\times \pi}$.

$m_e vr=n\times \frac {h}{2\times \pi}$ ....2

solving r from equation 2, we get that:

$r=n\times \frac {h}{2\times \pi\times m_e v}$

Putting in 1 , we get that:

$v=\frac{Ze^2}{2 \epsilon_0\times n\times h}$

Applying values for hydrogen atom,

Z = 1

Mass of the electron ($m_e$) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

$\epsilon_0$ = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

$v=\frac {2.185\times 10^6}{n}\ m/s$