Respuesta :
Answer:
[tex]v=\frac{Ze^2}{2 \epsilon_0\times n\times h}[/tex]
[tex]v=\frac {2.185\times 10^6}{n}\ m/s[/tex]
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,
[tex]\frac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\frac{v^2}{r}[/tex] ......1
Where, Z is the atomic number or the number of protons
r is the atomic radius
v is the velocity of the electron
[tex]m_e[/tex] 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 [tex]\frac {h}{2\times \pi}[/tex].
[tex]m_e vr=n\times \frac {h}{2\times \pi}[/tex] ....2
solving r from equation 2, we get that:
[tex]r=n\times \frac {h}{2\times \pi\times m_e v}[/tex]
Putting in 1 , we get that:
[tex]v=\frac{Ze^2}{2 \epsilon_0\times n\times h}[/tex]
Applying values for hydrogen atom,
Z = 1
Mass of the electron ([tex]m_e[/tex]) is 9.1093×10⁻³¹ kg
Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C
[tex]\epsilon_0[/tex] = 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:
[tex]v=\frac {2.185\times 10^6}{n}\ m/s[/tex]
