Answered: Calculate the energy separations in J,…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Calculate the energy separations in J, kJ/mol, eV (electron volts), and cm-1between the levels

a)n=2 and n=1

b)n=6 and n=5 of an electron in abox of length 1.50nm.

Expert Solution

Step 1

a) n=2 and n=1 of an electron in abox of length 1.50nm.

For energy separations, we can use the given below expression :

$E_{n + 1} - E_{n} = (2 n + 1) \frac{h^{2}}{8 m l^{2}} h e r e n = 1 h = P l a n c k' s c o n s \tan t = 6.626 \times 10^{- 34} J / s m = m a s s o f e l e c t r o n = 9.11 \times 10^{- 31} k g l = l e n g t h o f b o x = 1.50 n m = 1.5 \times 10^{- 9} m ∆ E = E_{2} - E_{1} = (2 + 1) \frac{6.626 \times 10^{- 34}}{8 \times (9.11 \times 10^{- 31}) \times {(1.5 \times 10^{- 9})}^{2}} = 8.04 \times 10^{- 20} J$

In kJ/mol,

$∆ E = 8.04 \times 10^{- 20} J \times \frac{6.02 \times 10^{23} p h o t o n s}{1 m o l p h o t o n s} \times \frac{1 k J}{10^{3} J} = 48.4 k J / m o l$

In eV,

$∆ E = 48.4 k J / m o l \times \frac{1 e V}{96.485 k J / m o l} = 0.502 e V$

In cm^-1,

$∆ E = 0.502 e V \times \frac{8065.5 c m^{- 1}}{1 e V} = 4048.88 c m^{- 1}$

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