   Chapter 19, Problem 16P

Chapter
Section
Textbook Problem

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 × 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?

To determine
The distance between the impact points formed on the photographic plate by singly charged ions of 235U and 238U .

Explanation

Given info: The ions in the beam emerge from the velocity selector at a speed of 3.00×105ms-1 . The magnetic field is 0.600T acting perpendicular to the velocity of the ions.

Explanation:

From the velocity selector particle only with velocity v=EB will emerge out. Hence the velocity of the ions is,

v=EB

• E is the electric field
• B is the magnetic field

In the deflection chamber the centripetal force is supplied by the magnetic field.

Hence,

qvB=mv2r

• q is the charge of the ion
• m is the mass of the ion
• r is the radius of the circular path of the ion

Re-arranging for the radius,

r=mvqB       (1)

The distance between the impact points will be the difference between the diameters of the circular path followed by the isotopes 235U and 238U .

d=2(r238r235)       (2)

• d is the distance between the impact points
• r238 is the radius of the path followed by 238U
• r235 is the radius of the path followed by 235U

Combining (1) and (2),

d=2vqB(m238m235)

• m238 is the mass of 238U
• m235 is the mass of 235U

Mass of 238U is 238u

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