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Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. What is the magnitude of the centripetal acceleration in m/sec2? (radius of the circle LaTeX: 5\times 10^{-11}5 × 10 − 11m; period of the motion LaTeX: 1.5 \times 10^{-16}1.5 × 10 − 16sec) Group of answer choices

User Malu
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1 Answer

9 votes

Answer:


A_c=87.73*10^(21)m/s

Step-by-step explanation:

From the question we are told that


r=5* 10^(-11)


T=1.5 * 10^(-16)

Generally the equation for velocity is mathematically given as


Velocity (v)=(2 \pi r)/(t)


V=(2 \pi (5*10^(-11)))/(1.5*10^(-16))


V=(2 \pi (5*10^(-11)))/(1.5*10^(-16))

Generally the equation for Centripetal acceleration is mathematically given as


A_c=(V^2)/(r)


A_c=((20.944*10^5))/(r5*10^(-11))


A_c=87.73*10^(21)m/s

User Sqeezer
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