Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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!
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.