{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1161_Physics ProblemsTechnical Physics

1161_Physics ProblemsTechnical Physics - 502*P41.30 Quantum...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
502 Quantum Mechanics *P41.30 The original tunneling probability is T e CL = 2 where C m U E = = × × × × = × 2 2 2 9 11 10 12 1 6 10 6 626 10 1 448 1 10 1 2 31 19 1 2 34 10 1 a f c h a f e j = π . . . . kg 20 J J s m . The photon energy is hf hc = = = λ 1 240 2 27 eV nm 546 nm eV . , to make the electron’s new kinetic energy 12 2 27 14 27 + = . . eV and its decay coefficient inside the barrier ′ = × × × × = × C 2 2 9 11 10 20 14 27 1 6 10 6 626 10 1 225 5 10 31 19 1 2 34 10 1 π . . . . . kg J J s m a f e j . Now the factor of increase in transmission probability is e e e e e C L CL L C C × × × = = = = 2 2 2 2 10 0 223 10 4.45 9 10 1 85 9 a f m m . . . Section 41.7 The Scanning Tunneling Microscope P41.31 With the wave function proportional to e CL , the transmission coefficient and the tunneling current are proportional to ψ 2 , to e CL 2 . Then, I I e e e 0 500 0 515 1 35 2 10 0 0 500 2 10 0 0 515 20 0 0 015 . . . . . . . . . nm nm nm nm nm nm a f a f b ga f b ga f a f = = = .
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}