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Atom 1 of mass 35uand atom 2 of mass 37u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (Fig. 28-12) and accelerated from rest through a potential difference V=7.3kV, each ion follows a circular path in a uniform magnetic field of magnitude B=0.50T. What is the distanceΔxbetween the points where the ions strike the detector?

Short Answer

Expert verified

The distance Δxbetween the points where the ions strike the detector is localid="1662915618019" Δx=8.2mm.

Step by step solution

01

Given

The Mass Of atom 1,m1=35u.

The Mass of atom 2, m2=37u.

Both the masses are slightly ionized with a charge of +e.

The potential difference is V=7.3kV.

The uniform magnetic field of magnitude B=0.50T.

02

Understanding the concept

By using the equations for the radius of the circular path and the speed of the ion, we can find the equation forΔmand by using this equation, we can find the distance Δxbetween the points where the ions strike the detector.

Formula:

The radius of the circular path, r=mv/qB

The speed,v=(2qV/m)

03

Calculate the distance Δx between the points where the ions strike the detector.

From equation 28-16, the radius of the circular path is

r=mvqB(1)

From equation 28-22, the speed of the ion is

v=2qVm

Substituting this value in equation (1),

localid="1662959036682" r=mqB2qVm

Simplifying further,

localid="1662959087996" r=2qVqB2

Since,

x=2r=22qVqB2=8mVqB2

Rearranging this equation for m, we get,

m=B2qx28V

From this we have

localid="1662959302850" Δm=B2q8V2xΔx

Substituting the value of x,

localid="1662959354410" width="186">Δm=B2q8V8mVqB2Δx

Δm=Bmq2VΔx

Thus, the distance between the spots made on the photographic plate is

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Most popular questions from this chapter

A proton, a deuteron (q=+e, m=2.0u), and an alpha particle (q=+2e, m=4.0u) are accelerated through the same potential difference and then enter the same region of uniform magnetic field, moving perpendicular to . What is the ratio of (a) the proton’s kinetic energy Kp to the alpha particle’s kinetic energy Ka and (b) the deuteron’s kinetic energy Kd to Ka? If the radius of the proton’s circular path is 10cm, what is the radius of (c) the deuteron’s path and (d) the alpha particle’s path

An electron that is moving through a uniform magnetic field has velocity v=(40km/s)+(35km/s)when it experiences a force F=(4.2 fN)+(4.8 fN)due to the magnetic field. If Bx=0, calculate the magnetic field B.

Bainbridge’s mass spectrometer, shown in Fig. 28-54, separates ions having the same velocity. The ions, after entering through slits, S1and S2, pass through a velocity selector composed of an electric field produced by the charged plates Pand P', and a magnetic field Bperpendicular to the electric field and the ion path. The ions that then pass undeviated through the crossedEand Bfields enter into a region where a second magnetic field Bexists, where they are made to follow circular paths. A photographic plate (or a modern detector) registers their arrival. Show that, for the ions, q/m=E/rBB,where ris the radius of the circular orbit.

Figure 28-27 shows the path of an electron that passes through two regions containing uniform magnetic fields of magnitudesB1and.B2

Its path in each region is a half-circle.

(a) Which field is stronger?

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(c) Is the time spent by the electron in theB1region greater than,

less than, or the same as the time spent in theB2region?

Question: At one instantv=(-2.00i^+4.00j^-6.00k^)m/s, is the velocity of a proton in a uniform magnetic fieldB=(2.00i^-4.00j^+8.00k^)mTAt that instant, what are (a) the magnetic force acting on the proton, in unit-vector notation, (b) the angle betweenv and F, and (c) the angle betweenv and B?

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