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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 6 - Problem 26e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 6 - Problem 26e

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# What is Hess’s law? Why is it useful? ISBN: 9780321809247 1

## Solution for problem 26E Chapter 6

Chemistry: A Molecular Approach | 3rd Edition

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Problem 26E

Problem 26E

What is Hess’s law? Why is it useful?

Step-by-Step Solution:
Step 1 of 3

Solution: Step1: a) The relation between wavelength() and frequency() of radiation is given by: = c / ------(1), where c is the velocity of light. In the above problem, given, wavelength of light = 532 nm = 532 x 10 m -9 [1 nm = 10 m]-9 = 5.32 x 10 m -7 8 Velocity of light, c = 3 x 10 m/sec Substituting the values in equation (1), we get, 8 -7 frequency() = ( 3 x 10 m/sec) / (5.32 x 10 m) 15 -1 = 0.564 x 10 sec . = 5.64 x 10 sec . -1 14 -1 Thus, frequency of the radiation is 5.64 x 10 sec . Step2: b) 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 , 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. Given, wavelength of the light = 532 nm -9 -9 = 532 x 10 m [1 nm = 10 m] = 5.32 x 10 m -7 -34 Planck's constant, h = 6.626 x 10 J.s Velocity of light, c = 3 x 10 m/sec Substituting the values in equation (2), we get, E = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (5.32 x 10 m) -7 = 3.74 x 10 J -19 -19 Thus, energy of one photon of the radiation is 3.74 x 10 J. Step3: c) The energy in the ground state can be calculated by using the following equation: E = -2.18 x 10 -18( ) -------(3) n2 For ground state, the value of n = 1. Substituting the value in equation (3), we get, E = -2.18 x 10 -18 1 ( 1 = -2.18 x 10 -18J Step4: -19 As per calculations above, the energy of the photon in the upper excited state = -3.74 x 10 J. The negative sign indicates emission of energy. So, the energy gap between the ground state and the excited state = E finalEinitial -18 -19 = (-2.18 x 10 J) - (-3.74 x 10 J) = (-2.18 x 10 -18J) - (-0.374 x 10 ) -18 -18 -18 = (-2.18 x 10 J) + (0.374 x 10 ) -18 = -1.81 x 10 J -18 Thus, the energy gap between the ground state and the excited state is 1.81 x 10 J. ------------------

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##### ISBN: 9780321809247

Since the solution to 26E from 6 chapter was answered, more than 355 students have viewed the full step-by-step answer. The answer to “What is Hess’s law? Why is it useful?” is broken down into a number of easy to follow steps, and 8 words. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. This full solution covers the following key subjects: state, Energy, light, laser, Ground. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. The full step-by-step solution to problem: 26E from chapter: 6 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM.

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