EBK ORGANIC CHEMISTRY
EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 15, Problem 15.6P
Interpretation Introduction

(a)

Interpretation:

The energy of light with a wavelength of 450nm is to be calculated.

Concept introduction:

The Plank-Einstein equation gives the measure of the energy of an electromagnetic wave. The expression for Plank-Einstein equation is given below.

E=hc/λ

Expert Solution
Check Mark

Answer to Problem 15.6P

The energy of light with a wavelength 450nm is 2.66×102kJmol1.

Explanation of Solution

The expression for the energy by Plank-Einstein equation is shown below.

E=hc/λ…(1)

Where,

E is the energy of the electromagnetic wave.

h is the Planks constant having value 6.626×1034Js.

c is the velocity of light having value 3×108ms1.

λ is the wavelength of light.

The wavelength is given as λ=450nm.

The relation between nm and m is given below.

1nm=109m

The conversion of 450nm into m is done as shown below.

450nm=450nm×109m1nm=450×109m

Now, substitute the values of h, c and λ in the equation (1) and solve for the energy E.

E=(6.626×1034Js)×(3×108ms1)(450×109m)=0.044173×1017J×103kJ1J=0.044173×1017×103kJ=44.173×1023kJ

The energy in kJmol1, is obtained by multiplying the energy calculated above (in kJ) by the total number of photons present in one mole of photon which is 6.022×1023mol1.

E=(44.173×1023kJ)×(6.022×1023mol1)=2.66×102kJmol1

Therefore, the energy of light for the given wavelength (450nm) is 2.66×102kJmol1.

Conclusion

The energy of light is 2.66×102kJmol1 that has wavelength 450nm.

Interpretation Introduction

(b)

Interpretation:

The energy of light with a wavelength 250nm is to be stated.

Concept introduction:

The Plank-Einstein equation gives the measure of the energy of an electromagnetic wave. The expression for Plank-Einstein equation is given below.

E=hc/λ

Expert Solution
Check Mark

Answer to Problem 15.6P

The energy of light with a wavelength 250nm is 4.79×102kJmol1.

Explanation of Solution

The expression for the energy by Plank-Einstein equation is shown below.

E=hc/λ…(1)

Where,

E is the energy of the electromagnetic wave.

h is the Planks constant having value 6.626×1034Js.

c is the velocity of light having value 3×108ms1.

λ is the wavelength of light.

The wavelength is given as λ=250nm.

The relation between nm and m is given below.

1nm=109m

The conversion of 250nm into m is done as shown below.

250nm=250nm×109m1nm=250×109m

Now, substitute the values of h, c and λ in the equation (1) and solve for the energy E.

E=(6.626×1034Js)×(3×108ms1)(250×109m)=0.0795×1017J×103kJ1J=0.0795×1017×103kJ=79.5×1023kJ

The energy in kJmol-1, is obtained by multiplying the energy calculated above (in kJ) by the total number of photons present in one mole of photon which is 6.022×1023mol-1.

E=(79.5×1023kJ)×(6.022×1023mol1)=4.79×102kJmol1

Therefore, the energy of light for the given wavelength (250nm) is 4.79×102kJmol1.

Conclusion

The energy of light is 4.79×102kJmol1 that has wavelength 250nm.

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Chapter 15 Solutions

EBK ORGANIC CHEMISTRY

Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42APCh. 15 - Prob. 15.43APCh. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50APCh. 15 - Prob. 15.51APCh. 15 - Prob. 15.52APCh. 15 - Prob. 15.53APCh. 15 - Prob. 15.54APCh. 15 - Prob. 15.55APCh. 15 - Prob. 15.56APCh. 15 - Prob. 15.57APCh. 15 - Prob. 15.58APCh. 15 - Prob. 15.59APCh. 15 - Prob. 15.60APCh. 15 - Prob. 15.61APCh. 15 - Prob. 15.62APCh. 15 - Prob. 15.63APCh. 15 - Prob. 15.64APCh. 15 - Prob. 15.65APCh. 15 - Prob. 15.66APCh. 15 - Prob. 15.67APCh. 15 - Prob. 15.68APCh. 15 - Prob. 15.69APCh. 15 - Prob. 15.70APCh. 15 - Prob. 15.71APCh. 15 - Prob. 15.72APCh. 15 - Prob. 15.73APCh. 15 - Prob. 15.74APCh. 15 - Prob. 15.75APCh. 15 - Prob. 15.76APCh. 15 - Prob. 15.77APCh. 15 - Prob. 15.78APCh. 15 - Prob. 15.79APCh. 15 - Prob. 15.80APCh. 15 - Prob. 15.81APCh. 15 - Prob. 15.82APCh. 15 - Prob. 15.83APCh. 15 - Prob. 15.84APCh. 15 - Prob. 15.85APCh. 15 - Prob. 15.86AP
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