Atomic Models - AP Physics 2

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Question

What is the speed of an electron in the first Bohr orbit in meters per second?

$k=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$h=6.63 * 10^{-34}J\cdot s$

Answer

To find the speed of the electron, use the following formula:

$v=\frac{2\pi ke^2}{nh}$

$k=\textup{Coulomb's constant}=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$n=\textup{first orbit}=1$

$h=\textup{Planks's constant}= 6.63 * 10^{-34}J\cdot s$

$e= \textup{electron charge}= 1.6 * 10^{-19}C$

Substitute all the knowns and solve for velocity.

$v=\frac{2\pi ke^2}{nh}=\frac{2\pi (8.99 * 10^9)(1.6 * 10^{-19})^2}{(1)(6.63 * 10^{-34})}$

$v=2.181 * 10^6 \frac{m}{s}$

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Question

What is the speed of an electron in the first Bohr orbit in meters per second?

$k=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$h=6.63 * 10^{-34}J\cdot s$

Answer

To find the speed of the electron, use the following formula:

$v=\frac{2\pi ke^2}{nh}$

$k=\textup{Coulomb's constant}=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$n=\textup{first orbit}=1$

$h=\textup{Planks's constant}= 6.63 * 10^{-34}J\cdot s$

$e= \textup{electron charge}= 1.6 * 10^{-19}C$

Substitute all the knowns and solve for velocity.

$v=\frac{2\pi ke^2}{nh}=\frac{2\pi (8.99 * 10^9)(1.6 * 10^{-19})^2}{(1)(6.63 * 10^{-34})}$

$v=2.181 * 10^6 \frac{m}{s}$

Compare your answer with the correct one above

Question

What is the speed of an electron in the first Bohr orbit in meters per second?

$k=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$h=6.63 * 10^{-34}J\cdot s$

Answer

To find the speed of the electron, use the following formula:

$v=\frac{2\pi ke^2}{nh}$

$k=\textup{Coulomb's constant}=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$n=\textup{first orbit}=1$

$h=\textup{Planks's constant}= 6.63 * 10^{-34}J\cdot s$

$e= \textup{electron charge}= 1.6 * 10^{-19}C$

Substitute all the knowns and solve for velocity.

$v=\frac{2\pi ke^2}{nh}=\frac{2\pi (8.99 * 10^9)(1.6 * 10^{-19})^2}{(1)(6.63 * 10^{-34})}$

$v=2.181 * 10^6 \frac{m}{s}$

Compare your answer with the correct one above

Question

What is the speed of an electron in the first Bohr orbit in meters per second?

$k=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$h=6.63 * 10^{-34}J\cdot s$

Answer

To find the speed of the electron, use the following formula:

$v=\frac{2\pi ke^2}{nh}$

$k=\textup{Coulomb's constant}=8.99* 10^9 \frac{N \cdot m^2}{C^2}$

$n=\textup{first orbit}=1$

$h=\textup{Planks's constant}= 6.63 * 10^{-34}J\cdot s$

$e= \textup{electron charge}= 1.6 * 10^{-19}C$

Substitute all the knowns and solve for velocity.

$v=\frac{2\pi ke^2}{nh}=\frac{2\pi (8.99 * 10^9)(1.6 * 10^{-19})^2}{(1)(6.63 * 10^{-34})}$

$v=2.181 * 10^6 \frac{m}{s}$

Compare your answer with the correct one above

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