{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ANS_EX_4 - Swieter p1 of 12 CHM1040 Answer to Exercise Set...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Swieter; p1 of 12 CHM1040 Answer to Exercise Set 4 QUESTIONS ON THE PROPERTIES OF LIGHT 1. m/s 10 x 2.9979 = c where c = or , = c 8 ν λ ν λ a) m 10 x 5.89 = s 10 x 5.09 m/s 10 x 2.9979 = 7 - 1 - 14 8 λ b) m 10 x 3.5 = s 10 x 8.6 m/s 10 x 2.9979 = 5 - 1 - 12 8 λ c) m 10 x 1.5 = s 10 x 2.0 m/s 10 x 2.9979 = 1 - 1 - 9 8 λ 2. s _ J 10 x 6.6262 = h where c _ h = E or _ h = E and c = or , _ = c 34 - λ ν λ ν ν λ a) s 10 x 1.7 = m 10 x 1.8 m/s 10 x 2.9979 = 1 - 10 2 - 8 ν J 10 x 1.1 = m 10 x 1.8 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = E 23 - 2 - 8 -34 b) s 10 x 6.25 = m 10 x 4.80 m/s 10 x 2.9979 = 1 - 15 8 - 8 ν J 10 x 4.14 = m 10 x 4.80 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = E 18 - 8 - 8 -34 c) s 10 x 9.83 = m 10 x 3.05 m/s 10 x 2.9979 = 1 - 14 7 - 8 ν J 10 x 6.51 = m 10 x 3.05 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = E 19 - 7 - 8 -34 3. E c _ h = or c _ h = E and h E = or , _ h = E λ λ ν ν a) s 10 x 1.05 = s _ J 10 x 6.6262 J 10 x 6.95 = 1 - 14 34 - -20 ν and, m 10 x 2.86 = J 10 x 6.95 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = 6 - 20 - 8 -34 λ b) s 10 x 8.99 = s _ J 10 x 6.6262 J 10 x 5.96 = 1 - 7 34 - -26 ν and, m 3.33 = J 10 x 5.96 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = 26 - 8 -34 λ c) s 10 x 3.02 = s _ J 10 x 6.6262 J 10 x 2.00 = 1 - 18 34 - -15 ν and,
Background image of page 1

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

View Full Document Right Arrow Icon
Swieter; p2 of 12 m 10 x 9.93 = J 10 x 2.00 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = 11 - 15 - 8 -34 λ 4. a) ... e therefor ; 1 = _ λ ν cm 10 x 6.67 = cm 10 m 1 x m 10 x 150 1 = _ 1 - 7 2 12 - ν This is in the X-ray region. b) ... e therefor ; _ _ c _ h = E ν J 10 x 7.946 = m 10 4000 x m/s 10 x 2.9979 x s J 10 x 6.6262 = E 20 - 2 - 8 34 - This is in the IR region. c) ... e therefor ; c = ν λ nm 10 x 2.00 = m 10 nm 1 x s 10 x 1.50 m/s 10 x 2.9979 = 2 9 - 1 - 15 8 λ This is in the UV region. d) ... e therefor ; c _ h = E λ J 10 x 2.0 = m 10 x 10 m/s 10 x 2.9979 x s _ J 10 x 6.6262 = E 24 - 2 - 8 -34 This is in the MW region. e) ... e therefor ; h E = ν s 10 x 1.5 = s _ J 10 x 6.6262 J 10 x 1.0 = 1 - 15 34 - -18 ν This is in the UV region. f) ... e therefor ; c = _ ν ν cm 10 x 4.00 = cm 10 m 1 x m/s 10 x 2.9979 s 10 x 1.20 = _ 1 - 3 2 8 1 - 14 ν This is in the IR region.
Background image of page 2
Swieter; p3 of 12 X:\~Natsci\ADJUNCT\bhaskar\pdf\1041\ANS_EX#4.DOC 5. ... e therefor ; c _ h = _ h = E λ ν J/photon 10 x 05 3. or J 10 x 05 3. = m 10 x 650 m/s 10 x 2.9979 x s _ J 10 x 6.6262 = E 19 - 6 19 - 6 9 - 8 -34 J/mol 10 x 1.84 = mol photons 10 x 6.022 x photon J 10 x 05 3. = mole photons # x photon E = E 5 23 -19 6 tot 6. J 10 x 3 1. = m 10 x 0.15 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = c _ h = E : rays X- 15 - 2 9 - 8 -34 λ J 10 x 98 1. = m 10 x 1.00 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = c _ h = E : Microwaves 22 - 6 3 - 8 -34 λ ! 10 x 6.7 = J 10 x 98 1. J 10 x 3 1. = Energy Microwave Energy ray X- 6 22 - 6 -15 2 7. ... e therefor ; c _ h = _ h = E photon λ ν J 10 x 97 3. = m 10 x 500 m/s 10 x 2.9979 x s _ J 10 x 6.6262 = E 19 - 3 9 - 8 -34 r q q F 2 2 1 8. J 10 x 31 4. = m 10 x 460 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = c _ h = E 19 - 8 9 - 8 -34 absorbed λ J 10 x 01 3. = m 10 x 660 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = c _ h = E 19 - 0 9 - 8 -34 emitted λ J 10 x 1.31 = J 10 x 01 3. - J 10 x 31 4. = E - E = E -19 -19 0 -19 8 emitted absorbed 9. a) J 10 x 8.6 = ) s 10 x (1.3 x s) _ J 10 x (6.6262 = _ h = E -19 -1 15 -34 ν b) J 10 x 1.99 = m 10 x 100 m/s) 10 x s)(2.9979 _ J 10 x (6.6262 = c _ h = E : nm) 100 _ ( UV 18 - 9 - 8 -34 λ Photoelectric effect is observed: Energy of UV photon exceeds threshold energy for Pt. b) J 10 x 1.99 = m 10 x 1.00 m/s) 10 x 9979 s)(2. J 10 x (6.6262 = c h = E : cm) 10 x 1.00 1 _ ( IR 20 - 5 - 8 -34 3 - λ Photoelectric effect is not observed: Energy of IR photon is less than threshold energy for Pt. c) Increasing the intensity increases the number of photons and not the energy per photon. Therefore, an increase in the UV intensity would increase the photoelectric effect ( current increases), while an increase in IR intensity would have no effect (more photons, but each with insufficient energy to remove an e - from Pt.
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}