Question

A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a speed of 3.00 3 105 m/s and enter a uniform magnetic field of 0.600 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by singly charged ions of 235U and 238U?

Answer 1

Δr = 15 m

Step-by-step explanation:

Let's use Newton's second law

F = m a

Force is magnetic

F = q v B sin θ

Where θ = 90 since it indicates that the field is perpendicular to the velocity. The acceleration is centripetal

a = v² / r

Let's replace

q v B = m v² / r

r = mv /q B

Let's look for the mass for each isotope

1 una = 1.660 10⁻²⁷ kg

235U m1 = 235 uma (1,660 10⁻²⁷kg / 1 unit) = 390.1 10⁻²⁷kg

238U m2 = 238 uma (1,660 10⁻²⁷ kg / 1 unit) = 395.1 10⁻²⁷ kf

In the exercises they indicate that ions have an unitary charge.

q = 1.6 10⁻¹⁹ C

Let's calculate the radius for each isotope

235U

r₁ = 390.1 10⁻²⁷ 3.00 10⁵ / ( 1.6 10⁻¹⁹ 0.600)

r₁ = 1170.3/0.96 10⁻³

r₁ = 1.219 m

238U

r₂ = 395.1 10⁻²⁷ 3.00 10⁵ /(1.6 10⁻¹⁹ 0.600)

r₂ = 1185.3/0.96 10⁻³

r₂ = 1.234 m

The distance between the two impact points is

Δr = r₂ -r₁

Δr = 1.234 - 1.219

Δr = 15 m