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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 6 - Problem 34e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 6 - Problem 34e

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# Convert between energy units:a. 231 cal to kJb. 132 × 104

ISBN: 9780321809247 1

## Solution for problem 34E Chapter 6

Chemistry: A Molecular Approach | 3rd Edition

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Problem 34E

Convert between energy units:

a. 231 cal to kJ                 b. 132 x 104 kJ to kcal

c. 4.99 x 103 kJ to kWh         d. 2.88 x 104 J to Cal

Step-by-Step Solution:
Step 1 of 3

Solution: The phenomenon of ejection of electrons from the surface of a metal when light of suitable frequency strikes on it, is called photoelectric effect. Step1: a) According to Planck’s quantum theory of radiation, the amount of energy associated with a quantum(in case of light, the quantum of energy is often called photon) of radiation is proportional to the frequency of light, E = h ---(1), where h is a universal constant called Planck’s constant. is the frequency and E is the energy of a photon. Or, E = hc/ -------(2), where, is the wavelength of light and c is the velocity. Step2: Here, given minimum energy required = 6.94 x 10 -19J Planck's constant, h = 6.626 x 10 J.s -34 Substituting the values in equation (1), we get, 6.94 x 10 -19J = (6.626 x 10 J.s) x -19 -34 = (6.94 x 10 J) / (6.626 x 10 J.s) = 1.05 x 10 s . -1 Thus, minimum frequency required to eject an electron is 1.05 x 10 s . 15 -1 Step3: b) From equation (2), = hc/E Here, Velocity of light, c = 3 x 10 m/sec 8 Energy, E = 6.94 x 10 -19J Substituting the values, we get, = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (6.94 x 10 -19J) -7 = 2.86 x 10 m. Thus, wavelength of the light = 2.86 x 10 m. -7 Step4: c) For each metal a minimum frequency of light called threshold frequency is required below which there is no ejection of electrons from the surface of the metal. The minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect is 1.05 x 10 s and the frequency of visible light ranges from (4 x 10 s ) to (8 x 10 s ) which is less14 -1 than the threshold frequency of titanium. Therefore, it is not possible to eject electrons from titanium metal using visible light. Step5: d) Wavelength of the irradiated light = 233 nm = 233 x 10 m -9 [1 nm = 10 m]-9 = 2.33 x 10 m -7 -34 Planck's constant, h = 6.626 x 10 J.s Velocity of light, c = 3 x 10 m/sec 8 Substituting the values in equation (2), we get, -34 8 -7 E = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (2.33 x 10 m) = 8.53 x 10 J -19 -19 Thus, energy of the irradiated radiation = 8.53 x 10 J Step6: Therefore, kinetic energy of the emitted electron = Energy of the radiation - minimum energy needed to eject an electron = (8.53 x 10 J) - (6.94 x 10 -19J) -19 = 1.59 x 10 J. Step7: -6 -6 e) Total energy of the light = 2.00 J = 2 x 10 J [1 J = 10 J] Energy required to eject an electron = 6.94 x 10 -19J Therefore, total number of electrons ejected = Total energy of light / energy required to eject 1 electron = (2 x 10 J) / (6.94 x 10 -19J) 13 = 0.288 x 10 12 = 2.88 x 10 . Thus, 2.88 x 10 electrons were ejected. --------------------------------------

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##### ISBN: 9780321809247

This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. The full step-by-step solution to problem: 34E from chapter: 6 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Since the solution to 34E from 6 chapter was answered, more than 690 students have viewed the full step-by-step answer. This full solution covers the following key subjects: light, electrons, Energy, titanium, emit. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. The answer to “?Convert between energy units: a. 231 cal to kJ b. 132 x 104 kJ to kcal c. 4.99 x 103 kJ to kWh d. 2.88 x 104 J to Cal” is broken down into a number of easy to follow steps, and 30 words. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247.

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