Problem 85 - Master Resource Book in JEE Main Physics

Question

The potential difference applied to an X-ray tube is \(5 \mathrm{kV}\) and current through it is \(3.2 \mathrm{~mA}\). Then, the number of electrons striking the target per second is (a) \(2 \times 10^{10}\) (b) \(3 \times 10^{18}\) (c) \(2 \times 10^{16}\) (d) \(5 \times 10^{15}\)

Step-by-Step Solution

Answer

The number of electrons striking the target per second is \(2 \times 10^{16}\). The correct option is (c).

Step 1: Understanding the Question

We need to find out how many electrons strike the target per second when a potential difference of \(5\, \mathrm{kV}\) is applied across an X-ray tube and the current through it is \(3.2\, \mathrm{mA}\).

Step 2: Convert Units

Convert the current from milliAmperes to Amperes: \(3.2\, \mathrm{mA} = 3.2 \times 10^{-3}\, \mathrm{A}\). No need to convert the potential difference for this problem.

Step 3: Formula for Number of Electrons

Use the formula relating current \(I\) and charge \(Q\): \(I = \frac{Q}{t}\). We find \(Q\) by multiplying \(I\) by \(t\). Since we want the number of electrons per second, \(t\) is 1 second.

Step 4: Calculate the Charge

Substitute \(I = 3.2 \times 10^{-3}\, \mathrm{A}\) and \(t = 1\, \mathrm{s}\): \(Q = I \cdot t = 3.2 \times 10^{-3}\, \mathrm{C}\).

Step 5: Number of Electrons

Use the charge of a single electron \(e = 1.6 \times 10^{-19}\, \mathrm{C}\) to find the number of electrons:\[ n = \frac{Q}{e} = \frac{3.2 \times 10^{-3}\, \mathrm{C}}{1.6 \times 10^{-19}\, \mathrm{C/electron}} = 2 \times 10^{16} \text{ electrons}. \]

Step 6: Select the Correct Option

Based on calculation, the number of electrons striking the target per second is \(2 \times 10^{16}\). The correct answer is (c) \(2 \times 10^{16}\).

Key Concepts

Electron Flow Calculation

Understanding electron flow in X-ray tubes involves figuring out how many electrons hit the target area per second.
In X-ray tubes, the potential difference across the tube causes electrons to accelerate and hit the target, generating X-rays. When you calculate electron flow, you need to consider both the current flowing through the tube and the charge carried by one electron.

• The current in the tube, measured in Amperes (A), tells you the total charge flowing per second.
• Knowing the charge of a single electron, you can calculate the total number of electrons striking the target each second.

This process won't only help you understand how X-ray images are produced but also enhance your grasp of electron interactions in electric circuits.

Current to Charge Conversion

In physics, we often relate current and charge using the formula: \( I = \frac{Q}{t} \), where \(I\) is the current in Amperes, \(Q\) is the total charge in Coulombs, and \(t\) is time in seconds. This relationship shows that current is the flow of charge per unit time. To find the total charge \(Q\), simply multiply current \(I\) by time \(t\).
For example, if the current through an X-ray tube is \(3.2\, \mathrm{mA}\), or \(3.2 \times 10^{-3}\, \mathrm{A}\), and we want to find the charge per second, we would multiply the current by 1 second:

• \(Q = 3.2 \times 10^{-3}\, \mathrm{C}\)
• This is the charge carried by all electrons passing through the tube each second.

Understanding this conversion is essential in solving problems involving electron numbers in X-ray technology.

Potential Difference in X-ray Tubes

An X-ray tube uses a high potential difference to accelerate electrons from the cathode to the anode.
This potential difference, measured in Volts, provides the energy needed to propel electrons at high speeds. The accelerated electrons then collide with the target material, producing X-rays.

• The potential difference transforms electrical energy into kinetic energy for the electrons.
• In this context, a \(5\,\mathrm{kV}\) potential means that each electron gains energy equivalent to \(5,000\) Joules per Coulomb of charge.

The concept of potential difference is crucial because it determines the energy—and thus the penetrating power—of the generated X-rays. A high potential difference results in high-energy X-rays, which are better for imaging denser parts of the body, such as bones.