Respuesta :
Answer:
a) v = 4.37 10⁶ m / s, speed is much less than c
b) v = 2.01 10⁸ m / s, this value is 67% of the speed of light, , for which relativistic corrections should be used
Explanation:
The bohr model for the hydrogen atom and dendroids is a classical model with a quantization of the angular momentum
let's start by using Newton's second law with the electric force
F = m a
Coulomb's law electric force
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case in an atom the number of protons is equal to the atomic number and there is only one electron
q₁ = Ze
q₂ = e
acceleration is centripetal
a = v² / r
we substitute
[tex]k \frac{Z e^2}{r^2} = m \frac{v^2}{r}[/tex]
v² = [tex]k \frac{Ze^2}{m r}[/tex]
quantization is imposed without justification in this model,
L = p x r = n [tex]\hbar[/tex]
\hbar= h /2π
if we consider circular orbits, the speed and position are perpendicular
m v r = n \hbar
r = [tex]\frac{n \hbar}{m v}[/tex]
we substitute
v² = [tex]k \frac{Z e^2}{m} \frac{m v}{n \hbar}[/tex]
v = [tex]k \frac{Z e^2 }{ n \hbar}[/tex]
let's apply this equation
\hbar= h / 2π
\hbar= 6.626 10-34 / 2π
\hbar= 1.05456 10⁻³⁴ J s
a) He1 ion, the atomic number of helium is 2
v = [tex]\frac{9 \ 10^9 \ 2 ( 1.6 \ 10^{-19})^2 }{n \ 1.0546 \ 10^{-34}}[/tex]
v =4.3695 10⁶ / n m / s
the ground state occurs for N = 1
v = 4.37 10⁶ m / s
the relationship of this value to the speed of light is
v / c = 4.37 10⁶/3 10⁸
v / c = 1.46 10⁻²
speed is much less than c
b) the uranium ion with atomic number Z = 92
v = [tex]\frac{9 \ 10^9 \ 92 ( 1.6 \ 10^{-19})^2 }{n \ 1.054 \ 10^{-34} }[/tex]
v = 2.01 10⁸ m / s
v/c = [tex]\frac{2.01 \ 10^8 }{3 \ 10^8}[/tex]
v/c = 0.67
this value is 67% of the speed of light, for atoms with a higher atomic number the effects are increasingly important, for which relativistic corrections should be used
