(a) In the hydrogen atom, an electron moves in a circular orbit about a proton, where the redius of the orbit is $5.29 \times 10^{-11}$ m. Find (i) the magnitude of the electric fo... (a) In the hydrogen atom, an electron moves in a circular orbit about a proton, where the redius of the orbit is $5.29 \times 10^{-11}$ 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 $5.00 \times 10^6$ 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 magnitude of the electric force between the electron and proton is given by Coulomb's law:

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

where $k = 8.987 \times 10^9 , \text{N m}^2/\text{C}^2$, $q_1 = q_p = +1.60 \times 10^{-19} , \text{C}$, $q_2 = q_e = -1.60 \times 10^{-19} , \text{C}$, and $r = 5.29 \times 10^{-11} , \text{m}$.

Plug in the values and solve for $F$:

$F = (8.987 \times 10^9) \frac{(1.60 \times 10^{-19})^2}{(5.29 \times 10^{-11})^2}$

2. (a)(ii) Calculate the electron speed

The electric force provides the centripetal force required for the circular motion of the electron:

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

where $m = 9.11 \times 10^{-31} , \text{kg}$ and $r = 5.29 \times 10^{-11} , \text{m}$. Solve for $v$:

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

Plug the force $F$ calculated in the previous step, along with the given values for $r$ and $m$.

3. (b)(i) Calculate the acceleration

The electric force on the electron is given by $F = qE$, where $E = 650 , \text{N/C}$ is the electric field. Using Newton's second law, $F = ma$, we can find the acceleration:

$a = \frac{qE}{m}$,

where $q=1.60 \times 10^{-19} , \text{C}$ and $m = 9.11 \times 10^{-31} , \text{kg}$.

4. (b)(ii) Calculate the time to reach the specified speed

Use the kinematic equation $v = v_0 + at$, where $v_0 = 0 , \text{m/s}$ (since the electron starts from rest), $v = 5.00 \times 10^6 , \text{m/s}$ and $a$ is calculated in the previous step. Solve for $t$:

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

5. (b)(iii) Calculate the distance traveled

Use the kinematic equation $x = v_0t + \frac{1}{2}at^2$. Since $v_0 = 0$, this simplifies to:

$x = \frac{1}{2}at^2$.

Plug in the values for $a$ and $t$ to find the distance $x$.