Question

In an atom electrons revolve around the nucleus along a path of radius 0.72 A making 9.4 x 10¹⁴ revolutions per second. The equivalent current is. (Given e = 1.6 x 10^-19 C)

• A. 1.4 A
• B. 1.2 A
• C. 1.8 A
• D. 1.5 A

Answer 1

The Correct Option is D

Consider an electron revolving around the nucleus of an atom. The electron's circular motion creates an equivalent electric current. We are given the following information:

• Radius of the circular path: \(r = 0.72\) Å (Angstroms) = \(0.72 \times 10^{-10}\) m
• Revolution frequency: \(f = 9.4 \times 10^{14}\) revolutions per second (Hz)
• Elementary charge: \(e = 1.6 \times 10^{-19}\) C

The equivalent current (\(I\)) is determined by the amount of charge passing a point in the orbit per unit time. Since each revolution carries a charge \(e\), and there are \(f\) revolutions per second, the total charge passing a point per second is \(I = e \times f\).

Therefore, the equivalent current is calculated as: 

\(I = e \times f\)

Substituting the given values:

\(I = (1.6 \times 10^{-19} \text{ C}) \times (9.4 \times 10^{14} \text{ Hz})\)

\(I = 1.6 \times 9.4 \times 10^{-19+14} \text{ A}\\  I = 15.04 \times 10^{-5} \text{ A}\\  I \approx 1.504 \text{ A}\)

Therefore, the equivalent current is approximately:

1.5 A

Answer 2

\( I = n \times q \times f \)

Given:

• n: \( 9.4 \times 10^{14} \, \text{revolutions per second} \)
• q: \( 1.6 \times 10^{-19} \, \text{C} \) (charge of each electron)
• f: \( 1 \) (since each electron completes one revolution per revolution)

Step 1: Plug in the given values into the formula for current:

\( I = (9.4 \times 10^{14}) \times (1.6 \times 10^{-19}) \times 1 \)

Step 2: Perform the calculation:

\( I = 1.504 \, \text{A} \)

Therefore, the equivalent current is: \( I = 1.5 \, \text{A} \), rounded to one decimal place. The closest option is (D) 1.5 A.

Answer 3

We are given the following information: 

• Radius of the path, \( r = 0.72 \, \text{Å} = 0.72 \times 10^{-10} \, \text{m} \) (This information is not directly needed to calculate the current).
• Frequency of revolution, \( f = 9.4 \times 10^{14} \, \text{revolutions/second} = 9.4 \times 10^{14} \, \text{Hz} \)
• Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

An electron revolving in a circular path constitutes an electric current. The magnitude of the equivalent current \( I \) is the total charge passing through any point on the path per unit time.

In one second, the electron makes \( f \) revolutions. Therefore, the total charge passing a point in one second is the number of revolutions multiplied by the charge of the electron.

The formula for the equivalent current is:

\[ I = \text{charge} \times \text{frequency} \]

\[ I = e \times f \]

Substituting the given values:

\[ I = (1.6 \times 10^{-19} \, \text{C}) \times (9.4 \times 10^{14} \, \text{s}^{-1}) \]

\[ I = (1.6 \times 9.4) \times (10^{-19} \times 10^{14}) \, \text{A} \]

\[ I = 15.04 \times 10^{-5} \, \text{A} \]

\[ I = 1.504 \times 10^{-4} \, \text{A} \]

This result (\( 1.504 \times 10^{-4} \, \text{A} \)) does not match any of the given options (1.4 A, 1.2 A, 1.8 A, 1.5 A), which are significantly larger.

It is highly probable that there is a typographical error in the frequency given in the question. Let's assume the frequency was intended to be \( 9.4 \times 10^{18} \, \text{Hz} \) to see if it matches any option.

Assuming \( f = 9.4 \times 10^{18} \, \text{Hz} \):

\[ I = e \times f \]

\[ I = (1.6 \times 10^{-19} \, \text{C}) \times (9.4 \times 10^{18} \, \text{s}^{-1}) \]

\[ I = (1.6 \times 9.4) \times (10^{-19} \times 10^{18}) \, \text{A} \]

\[ I = 15.04 \times 10^{-1} \, \text{A} \]

\[ I = 1.504 \, \text{A} \]

This value \( 1.504 \, \text{A} \) is very close to the option 1.5 A.

Therefore, assuming the frequency was intended to be \( 9.4 \times 10^{18} \, \text{Hz} \), the equivalent current is approximately 1.5 A.