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Sagot :
Answer:
5.3×10⁻¹¹ m
Step-by-step explanation:
Applying,
v = 2πr/t................ Equation 1
Where v = speed of the electron in the orbit, r = radius of the orbit, t = time, π = pie
make r the subject of the equation
r = vt/2π................ Equation 2
From the question,
Given: v = 2.2×10⁶ m/s, t = 1.5×10⁻¹⁶ s
Constant: π = 3.14
Susbtitute these values into equation 2
r = (2.2×10⁶×1.5×10⁻¹⁶)/(3.14×2)
r = (3.3×10⁻¹⁰)/6.28
r = 5.3×10⁻¹¹ m
Hence the radius of the orbit is 5.3×10⁻¹¹ m
Answer:
The radius of the orbit is: [tex]R=0.525 \AA[/tex]
Step-by-step explanation:
The period of a circular motion is the time to complete one orbit, then:
[tex]T=1.5*10^{-16}s[/tex]
Now, let's recall the tangential speed can be written as:
[tex]v=\frac{2\pi R}{T}[/tex]
R is the radius of the motion.
Let's solve the above equation for R.
[tex]R=\frac{vT}{2 \pi}[/tex]
[tex]R=\frac{2.2*10^{6}1.5*10^{-16}}{2 \pi}[/tex]
[tex]R=\frac{2.2*10^{6}1.5*10^{-16}}{2 \pi}[/tex]
[tex]R=5.25*10^{-11} \: m[/tex]
Therefore, the radius of the orbit is: [tex]R=0.525 \AA[/tex]
I hope it helps you!
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