s04x3aan - = =-m r L d d = 2.0 m b) All of the power from...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
NAME PHYS-1200 PHYSICS II QUIZ 3 APRIL 23, 2004 SOLUTIONS PART A. 1. D 7. A 2. E 8. A 3. D 4. E 5. A 6. D PART B . 1. ___ The maximum kinetic energy of the emitted electrons (Depends on photon frequency and work function only) _ _ The maximum photoelectric current ___ The stopping potential (Depends on photon frequency and work function only) ___ The cutoff frequency (Depends on work function only) 2. According to the hint Probability = x dx x x = 2 2 2 2 | | | | 2 1 ψ . Then, Probability = = = 2 sin 02 . 0 ) pm 0 . 2 ( pm 200 ) pm 50 ( 2 sin pm 200 2 2 sin 2 2 2 2 π x L x L Probability = 0.020 = 2.0% 3. _F _ Both substances have the same number density of charge carriers at room temperature. _T _ At room temperature, germanium has a greater number density of charge carriers than silicon. _F _ Both substances have a greater number density of conduction electrons than holes. _T _ For each substance, the number density of electrons equals that of holes.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
NAME PART C. 1. a) d λ θ 22 . 1 = , and L r m = so d L r m 22 . 1 = , and m 160 m) 10 82 . 3 ( m) 10 90 . 6 ( 22 . 1 22 . 1 8 7 × ×
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = =-m r L d d = 2.0 m b) All of the power from the laser falls in a circle of radius r m on the moon, so the average intensity is given by 2 6 2 m) (160 W 10 . 2 = = = m r P A P I I = 25 W/m c) BOTH The power of the laser determines the power that reaches the moon, and the wavelength determines the area over which it is spread. 2. a) METAL b) E F is the energy of the highest occupied state, measured from the bottom of the conduction band. E F = 24 eV 20 eV E F = 4 eV c) is the least energy needed to free an electron from the material. It equals the difference between the top of the wells and the Fermi energy. = 32 eV 24 eV = 8 eV d) m 10 150 m/s) 10 . 3 s)( J 10 63 . 6 ( 9 8 34-- = = = hc hf E ph E ph = 1.33 10-18 J = 8.3 eV e) YES The energy of the photon is greater than the work function of the material....
View Full Document

Page1 / 2

s04x3aan - = =-m r L d d = 2.0 m b) All of the power from...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online