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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

Sagot :

Okpalawalter8idn

Answer:

[tex]A_c=87.73*10^{21}m/s[/tex]

Explanation:

From the question we are told that

[tex]r=5\times 10^{-11}[/tex]

[tex]T=1.5 \times 10^{-16}[/tex]

Generally the equation for velocity is mathematically given as

[tex]Velocity (v)=\frac{2 \pi r}{t}[/tex]

[tex]V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}[/tex]

[tex]V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}[/tex]

Generally the equation for Centripetal acceleration is mathematically given as

[tex]A_c=\frac{V^2}{r}[/tex]

[tex]A_c=(\frac{20.944*10^5)}{r5*10^{-11}}[/tex]

[tex]A_c=87.73*10^{21}m/s[/tex]

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