Question icon
Grade 11General Physics

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.

Profile image of Jayant Kumar
12 Years agoGrade 11
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer11 Months ago

To determine the charge-to-mass ratio (e/m) of the electrons in the given scenario, we can use the information provided about the electron gun, the magnetic field, and the radius of the circular path of the electrons. Let's break this down step by step.

Understanding the Basics

The charge-to-mass ratio (e/m) is a fundamental property of charged particles, such as electrons. It can be calculated using the formula:

e/m = (2V) / (B²r²)

Where:

  • V = potential difference (in volts)
  • B = magnetic field strength (in teslas)
  • r = radius of the circular path (in meters)

Gathering the Data

From the problem statement, we have:

  • Potential difference, V = 100 V
  • Magnetic field strength, B = 2.83 × 10-4 T
  • Radius of the circular path, r = 12.0 cm = 0.12 m

Plugging in the Values

Now, we can substitute these values into the formula to find e/m:

e/m = (2 × 100 V) / ((2.83 × 10-4 T)² × (0.12 m)²)

Calculating Step-by-Step

First, calculate the denominator:

B² = (2.83 × 10-4)² = 8.0089 × 10-8

r² = (0.12)² = 0.0144 m²

Now, multiply B² by r²:

B²r² = (8.0089 × 10-8 T²) × (0.0144 m²) = 1.1533 × 10-9 T²·m²

Next, calculate the numerator:

2V = 2 × 100 V = 200 V

Now, we can find e/m:

e/m = 200 V / (1.1533 × 10-9 T²·m²)

Calculating this gives:

e/m ≈ 1.733 × 1011 C/kg

Final Result

Thus, the charge-to-mass ratio (e/m) of the electrons in this experiment is approximately 1.733 × 1011 C/kg. This value is consistent with the known charge-to-mass ratio of electrons, which is about 1.76 × 1011 C/kg, confirming the accuracy of our calculations.