Question

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3 x 10^-11 m with a speed of 2.2 x 10^6 m/s.
If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude of the electric field that the electron produces at the location of the nucleus (treated as a point).

Answer 1

The magnitude of the electric field is about 5.1 × 10¹¹ N/C

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Further explanation

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

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There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

$\boxed {F = k \frac{Q_1Q_2}{R^2} }$

$\boxed {E = F \div q = k \frac{Q}{R^2} }$

F = electric force (N)

E = electric field strength (N/C)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

Let's tackle the problem now !

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Given:

radius of orbit = R = 5.3 × 10⁻¹¹ m

speed of electron = v = 2.2 × 10⁶ m/s

charge of electron = Q = 1.6 × 10⁻¹⁹ C

Asked:

magnitude of the electric field = E = ?

Solution:

$E = k \frac{Q}{R^2}$

$E = 9 \times 10^9 \times \frac{1.6 \times 10^{-19}}{(5.3 \times 10^{-11})^2}$

$E \approx 5.1 \times 10^{11} \texttt{ N/C}$

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Learn more

• The three resistors : brainly.com/question/9503202
• A series circuit : brainly.com/question/1518810
• Compare and contrast a series and parallel circuit : brainly.com/question/539204

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Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

Answer 2

Answer:

The magnitude of the electric field is $0.513\times10^{12}\ N/C$.

Explanation:

Given that,

Radius $r= 5.3\times10^{-11}\ m$

Speed $v =2.2\times10^{6}\ m/s$

We need to calculate the electric field

Using formula of the electric field

$E=\dfrac{kQ}{r^2}$

Where, Q = charge

r = radius

Put the value into the formula

$E=\dfrac{9\times10^{9}\times1.6\times10^{-19}}{(5.3\times10^{-11})^2}$

$E=0.513\times10^{12}\ N/C$

Hence, The magnitude of the electric field is $0.513\times10^{12}\ N/C$.