Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 28, Problem 66P

(a)

To determine

The speed of the selected neutrons escaping the reactor.

(a)

Expert Solution
Check Mark

Answer to Problem 66P

The speed of the neutrons is 2.21×1034 m.

Explanation of Solution

The time interval between the two shutters is 13.0ms and the distance between the first and second shutter is 16.4 m.

Write the expression for speed.

v=xt                                                                          (I)

Here, v is the speed of the neutron, x is the distance travelled and t is the time of travel.

Substituting  13.0ms for t and 16.4 m for x in (I) to find v

v=16.4 m13.0ms=16.4 m13.0×103s=1.26×103ms1=1.26 kms1

Thus, the speed of the neutrons is 1.26 kms1.

(b)

To determine

The de Broglie wavelength of the neutrons.

(b)

Expert Solution
Check Mark

Answer to Problem 66P

The de Broglie wavelength of the neutrons is 314 pm.

Explanation of Solution

The velocity of the neutrons is 1.26×103ms1.

Write the expression for de Broglie wavelength

λ=hp                                           (II)

Here, λ is the de Broglie wavelength, h is the Planck’s constant  and p is the momentum of the bullet.

Substitute p=mv in (I)

λ=hmv                                           (III)

Here, m is the mass of the neutrons and v is its velocity.

Substitute 1.675×1027 kg for m, 1.26×103ms1 for v and 6.626×1034 Js for h in (III) to find λ

λ=6.626×1034 kgm2s11.675×1027 kg×1.26×103ms1=314×1012 m=314 pm

Thus, the de Broglie wavelength of the neutrons is 314 pm.

(c)

To determine

The range of de Broglie wavelength.

(c)

Expert Solution
Check Mark

Answer to Problem 66P

The range of de Broglie wavelength is 303 pm to 324 pm.

Explanation of Solution

The slowest neutron takes longer time to get through the shutters. It is given by the sum of the time interval between the two shutters and the time the two shutters are open i.e.

ts=t+t1                                                                  (IV)

Here, ts is the time of travel of the slowest neutron and t1 is the opening time of the shutter.

However, the fastest neutron takes lesser time and it is given by the difference of the time interval between the two shutters and the time the two shutters are open i.e.

tf=tt1                                                                (V)

Here, tf is the time of travel of the fastest neutron.

Substitute 0.45 ms for t1  and 13.0ms for t in (IV)

ts=0.45 ms+13.0ms=13.45ms

Substitute 13.45ms for t and 16.4 m for x in (I) to find v

v=16.4 m13.45ms=16.4 m13.45×103s=1220ms1

Thus, the speed of the slowest neutrons is 1220ms1.

Substitute 1.675×1027 kg for m, 1.26×103ms1 for v and 6.626×1034 Js for h in (III) to find λ

λ=6.626×1034 kgm2s11.675×1027 kg×1220ms1=324×1012 m=324 pm

Thus, the de Broglie wavelength of the fastest neutrons is 324 pm.

Substitute 0.45 ms for t1  and 13.0ms for t in (V)

tf=0.45 ms13.0ms=12.55ms

Substitute 12.55ms for t and 16.4 m for x in (I) to find v

v=16.4 m12.55ms=16.4 m12.55×103s=1307ms1

Thus, the speed of the fastest neutrons is 1307ms1.

Substitute 1.675×1027 kg for m, 1307ms1 for v and 6.626×1034 Js for h in (III) to find λ

λ=6.626×1034 kgm2s11.675×1027 kg×1307ms1=303×1012 m=303 pm

Thus, the de Broglie wavelength of the slowest neutrons is 303 pm.

Thus, the range of de Broglie wavelength is 303 pm to 324 pm.

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Chapter 28 Solutions

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