1126_PartUniversity Physics Solution

1126_PartUniversity Physics Solution - Molecules and...

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Molecules and Condensed Matter 42-7 Then can calculate I for 10.0 mV: V = 19 3 23 (1.602 10 C)(10.0 10 V) 0.3867 (1.381 10 J/K)(300 K) eV kT −− ×× == × / 0.3867 s ( 1) (11.77 mA)( 5.56 mA eV kT IIe e =− = = (b) eV kT has the same magnitude as in part (a) but not V is negative so eV kT is negative. 15.0 mV : 0.5800 eV V kT and / 0.5800 s ( (11.77 mA)( 5.18 mA eV kT e = = 10.0 mV : 0.3867 eV V kT and / 0.3867 s ( (11.77 mA)( 3.77 mA eV kT e = = EVALUATE: There is a directional asymmetry in the current, with a forward-bias voltage producing more current than a reverse-bias voltage of the same magnitude, but the voltage is small enough for the asymmetry not be pronounced. Compare to Example 42.11, where more extreme voltages are considered. 42.32. (a) Solving Eq.(42.23) for the voltage as a function of current, S 40.0 mA ln 1 ln 1 0.0645 V. 3.60 mA kT I kT V eI e ⎛⎞ =+ = + = ⎜⎟ ⎝⎠ (b) From part (a), the quantity 12.11 eV kT e = , so far a reverse-bias voltage of the same magnitude, () SS 1 1 1 3.30 mA 12.11 eV kT I = . 42.33. IDENTIFY: During the transition, the molecule emits a photon of light having energy equal to the energy difference between the two vibrational states of the molecule. SET UP: The vibrational energy is r 11 22 n k En n m ω ⎛⎞⎛⎞ ⎜⎟⎜⎟ ⎝⎠⎝⎠ . EXECUTE: (a) The energy difference between two adjacent energy states is r k E m Δ= = , and this is the energy of the photon, so Δ E = hc/ λ . Equating these two expressions for Δ E and solving for k , we have 2 r E km Δ = 2 HO mm E Δ + = , and using / 2 Eh c c λπ Δ with the appropriate numbers gives us ( ) 2 27 26 8 27 26 6 1.67 10 kg 2.656 10 kg 2 3.00 10 m/s 1.67 10 kg + 2.656 10 kg 2.39 10 m k π × = × = 977 N/m (b) r 2 k f mk ππ + = . Substituting the appropriate numbers gives us ( ) 27 26 27 26 14 1.67 10 kg 2.656 10 kg 1 1.67 10 kg + 2.656 10 kg 1.25 10 Hz 29 7 7 N / m f × EVALUATE: The frequency is close to, but not quite in, the visible range. 42.34. 2 48 2 2 2 7.14 10 kg m 2 h I E π c Δ = . 42.35. IDENTIFY and SET UP: Eq.(21.14) gives the electric dipole moment as , pq d = where the dipole consists of charges q ± separated by distance d . EXECUTE: (a) Point charges e + and e separated by distance d , so 19 9 29 ( 1 . 6 0 21 0 C ) ( 0 . 2 41 0 m )3 . 81 Cm pe d −−− × × (b) d = so 29 19 9 3.0 10 C m 1.3 10 C 0.24 10 m p q d ×⋅ = × × (c) 19 19 1.3 10 C 0.81 1.602 10 C q e × ×
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42-8 Chapter 42 (d) 30 21 9 1.5 10 C m 9.37 10 C 0.16 10 m p q d ×⋅ == = × × 21 19 9.37 10 C 0.058 1.602 10 C q e × × EVALUATE: The fractional ionic character for the bond in HI is much less than the fractional ionic character for the bond in NaCl. The bond in HI is mostly covalent and not very ionic. 42.36. The electrical potential energy is 5.13 eV, U =− and 2 10 0 1 2.8 10 m. 4 e r π U = × P 42.37. (a) IDENTIFY: + (Na) (Cl) (Na ) (Cl ) ( ). EEE EU r += + + Solving for () Ur gives + ( ) [ (Na ) (Na)] [ (Cl) (Cl )]. E E E E + SET UP: + [ (Na ) (Na)] EE is the ionization energy of Na, the energy required to remove one electron, and is equal to 5.1 eV.
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1126_PartUniversity Physics Solution - Molecules and...

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