a) In the hydrogen atom, an electron moves in a circular orbit about a proton, where the radius of the orbit is 5.29 x 10^-11 m. Find: i) the magnitude of the electric force exerte... a) In the hydrogen atom, an electron moves in a circular orbit about a proton, where the radius of the orbit is 5.29 x 10^-11 m. Find: i) the magnitude of the electric force exerted on the electron and ii) the speed of the electron. b) An electron accelerates from rest in a uniform electric field of 650 N/C. At one later moment, its speed is 5.00 x 10^6 m/s. i) Find the acceleration of the electron. ii) How long does the electron take to reach that speed? iii) How far does it move in this time interval? Read more

Steps to Solve

1. (a)(i) Calculate the electric force

The electric force between the electron and proton is given by Coulomb's law:

$F = k \frac{|q_1 q_2|}{r^2}$

where $k = 8.9875 \times 10^9 , \text{N m}^2/\text{C}^2$ is Coulomb's constant, $q_1$ and $q_2$ are the charges of the proton and electron respectively, and $r$ is the distance between them. Since the charge of the proton is $+e$ and the charge of the electron is $-e$, we have $|q_1 q_2| = e^2$. Plugging in the values:

$F = (8.9875 \times 10^9) \frac{(1.60 \times 10^{-19})^2}{(5.29 \times 10^{-11})^2}$

2. Simplify and find the electric force

$F = (8.9875 \times 10^9) \frac{2.56 \times 10^{-38}}{2.79841 \times 10^{-21}}$

$F = \frac{2.3008 \times 10^{-28}}{2.79841 \times 10^{-21}}$

$F \approx 8.22 \times 10^{-8} , \text{N}$

3. (a)(ii) Find the speed of the electron

Since the electric force provides the centripetal force for the electron's circular motion, we can equate the electric force to the centripetal force:

$F = \frac{mv^2}{r}$

Where $m$ is the mass of the electron and $v$ is its speed. Rearrange to solve for $v$:

$v = \sqrt{\frac{Fr}{m}}$

4. Plug in values and find the electron speed

$v = \sqrt{\frac{(8.22 \times 10^{-8})(5.29 \times 10^{-11})}{9.11 \times 10^{-31}}}$

$v = \sqrt{\frac{4.348 \times 10^{-18}}{9.11 \times 10^{-31}}}$

$v = \sqrt{4.773 \times 10^{12}}$

$v \approx 2.18 \times 10^6 , \text{m/s}$

5. (b)(i) Find the acceleration of the electron

The force on the electron due to the electric field is $F = qE$, where $q$ is the charge of the electron and $E$ is the electric field strength. Using Newton's second law, $F = ma$, therefore $a = \frac{F}{m} = \frac{qE}{m}$.

$a = \frac{(1.60 \times 10^{-19})(650)}{9.11 \times 10^{-31}}$

6. Simplify and find the acceleration

$a = \frac{1.04 \times 10^{-16}}{9.11 \times 10^{-31}}$

$a \approx 1.14 \times 10^{14} , \text{m/s}^2$

7. (b)(ii) Find the time to reach the speed

We can use the equation $v = v_0 + at$, where $v_0$ is the initial velocity (0 in this case), $v$ is the final velocity, $a$ is the acceleration, and $t$ is the time. Solving for $t$:

$t = \frac{v - v_0}{a} = \frac{v}{a}$

$t = \frac{5.00 \times 10^6}{1.14 \times 10^{14}}$

8. Simplify and find the time

$t \approx 4.39 \times 10^{-8} , \text{s}$

9. (b)(iii) Find the distance traveled

We can use the equation $d = v_0t + \frac{1}{2}at^2$. Since $v_0 = 0$, we have:

$d = \frac{1}{2}at^2$

$d = \frac{1}{2}(1.14 \times 10^{14})(4.39 \times 10^{-8})^2$

10. Simplify and find the distance

$d = \frac{1}{2}(1.14 \times 10^{14})(1.927 \times 10^{-15})$

$d = (0.57 \times 10^{14})(1.927 \times 10^{-15})$

$d \approx 0.11 , \text{m}$