General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Question
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Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than 1MeV.

(a)

Expert Solution
Check Mark

Answer to Problem 48E

The energy of electron is 3.8125×108eV which is greater than 1MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

  ΔxΔp                                                                                                

Here, Δx is the change in position, Δp is the change in momentum and is the reduced Planck’s constant.

Rearrange the above expression for Δp.

    ΔpΔx                                                                                                

Write the expression for energy.

    E=(Δp)22m        (I)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute Δx for Δp in equation (I).

    E(Δx)22m=22m(Δx)2

Substitute 1.055×1034Js for , 9.1×1031kg for m and 1014m for Δx in above equation.

    E(1.055×1034Js)22(9.1×1031kg)(1014m)2=(1.055×1034Js)22(9.1×1031kg)(1014m)2(1kgm2s21J)=6.1×1011J(1eV1.6×1019J)=3.8125×108eV

Thus, the energy of electron is 3.8125×108eV which is greater than 1MeV.

(b)

To determine

To see that the energy of neutron is greater than 1MeV or not.

(b)

Expert Solution
Check Mark

Answer to Problem 48E

The energy of neutron is 2.06×105eV which is less than 1MeV.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

  ΔxΔp                                                                                                

Here, Δx is the change in position, Δp is the change in momentum and is the reduced Planck’s constant.

Rearrange the above expression for Δp.

    ΔpΔx                                                                                                

Write the expression for energy.

    E=(Δp)22m        (II)

Here, E is the energy, Δp is change in momentum and m is the mass.

Conclusion:

Substitute Δx for Δp in equation (II).

    E(Δx)22m=22m(Δx)2

Substitute 1.055×1034Js for , 1.67×1027kg for m and 1014m for Δx in above equation.

    E(1.055×1034Js)22(1.67×1027kg)(1014m)2=(1.055×1034Js)22(1.67×1027kg)(1014m)2(1kgm2s21J)=3.3×1014J(1eV1.6×1019J)=2.06×105eV

Thus, the energy of neutron is 2.06×105eV which is less than 1MeV.

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