Amass 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?

  Δr =  15 m

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

it's a b and

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.