6) An electron in a box Goal: Understanding Heisenberg's uncertainty principle Suppose an electron is confined to a box that is about 0.50 x 10-10 m on a side. If this distance is taken as the uncertainty in the position of the electron, (a) Calculate the corresponding minimum uncertainty in the momentum. 2.1 x1024 (b) Because the electron is confined to a stationary box, its average momentum is zero. The magnitude of the electron's momentum is nonzero, however. Assuming the magnitude of the electron's momentum is the same as its uncertainty in momentum, calculate the corresponding kinetic energy. 2.4×10-3 (1) a) 2.18 eV b) 1.16 x 10-27 kg m/s (2) a) 5.5 x 10 Hz b) 1.59 x 10-19 J (3) a) 6.55 x 10-10 b) 3.04 x 10-16 J c) 2 x 10-18 J (4) 2080 m/s (5) a) 6.64 x 10-5 m b) 1.72 x 10-34 m x 104 kg m/s b) 2.4 x 10-18J (6) a) 2.1

Physics for Scientists and Engineers with Modern Physics
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Chapter39: Introduction To Quantum Physics
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11.6.2
6) An electron in a box
Goal: Understanding Heisenberg's uncertainty principle
Suppose an electron is confined to a box that is about 0.50 x 10-10 m on a side. If this distance is
taken as the uncertainty in the position of the electron,
(a) Calculate the corresponding minimum uncertainty in the momentum. 2.1 x1024
(b) Because the electron is confined to a stationary box, its average momentum is zero. The
magnitude of the electron's momentum is nonzero, however. Assuming the magnitude of the
electron's momentum is the same as its uncertainty in momentum, calculate the corresponding
kinetic energy. 2.4×10-3
(1) a) 2.18 eV b) 1.16 x 10-27 kg m/s (2) a) 5.5 x 10 Hz b) 1.59 x 10-19 J (3) a) 6.55 x 10-10
b) 3.04 x 10-16 J c) 2 x 10-18 J (4) 2080 m/s (5) a) 6.64 x 10-5 m b) 1.72 x 10-34 m
x 104 kg m/s b) 2.4 x 10-18J
(6) a) 2.1
Transcribed Image Text:6) An electron in a box Goal: Understanding Heisenberg's uncertainty principle Suppose an electron is confined to a box that is about 0.50 x 10-10 m on a side. If this distance is taken as the uncertainty in the position of the electron, (a) Calculate the corresponding minimum uncertainty in the momentum. 2.1 x1024 (b) Because the electron is confined to a stationary box, its average momentum is zero. The magnitude of the electron's momentum is nonzero, however. Assuming the magnitude of the electron's momentum is the same as its uncertainty in momentum, calculate the corresponding kinetic energy. 2.4×10-3 (1) a) 2.18 eV b) 1.16 x 10-27 kg m/s (2) a) 5.5 x 10 Hz b) 1.59 x 10-19 J (3) a) 6.55 x 10-10 b) 3.04 x 10-16 J c) 2 x 10-18 J (4) 2080 m/s (5) a) 6.64 x 10-5 m b) 1.72 x 10-34 m x 104 kg m/s b) 2.4 x 10-18J (6) a) 2.1
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