Concept explainers
(i)
Interpretation: Under the given conditions, the energy of quantum has to be calculated.
Concept introduction: The quantum of energy is called ass photon. The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons. Photon is the fundamental particles of visible light. The energy of quantum is directly proportional to the frequency of radiation.
(i)
Answer to Problem 7A.4BE
The energy of quantum in joule and kilo joule per mole are 2.65×10-19 J_ and 159.6 kJ mol-1_ respectively.
Explanation of Solution
The energy of quantum is directly proportional to the frequency of radiation
Therefore,
E=hν (1)
Where,
- E is the energy of quantum.
- ν is the frequency of radiation.
- h is the Planck’s constant (6.626×10−34 J s)
The frequency in terms of time period is expressed as,
ν=1T (2)
Where,
- T is the time period of electronic oscillation.
Substitute the value of ν in equation (1).
E=h×1T (3)
It is given that,
The time period is 2.50 fs.
The conversion of fs to s is done as,
1 fs=10−15 s
Therefore, the conversion of 2.50 fs to s is done as,
2.50 fs=2.50×10−15 s
Substitute the value of h and T in equation (3).
E=6.626×10−34 J s×12.50×10−15 s=2.65×10-19 J_
Per mole energy (ENA) is calculated by using the formula,
ENA=NA×E (4)
Where,
- NA is Avogadro’s number (6.023×1023).
Substitute the value of NA and E in equation (4).
ENA=6.023×1023 mol−1×2.65×10−19 J =15.96×104 J mol−1=159.6×103 J mol−1
The conversion of J mol−1 to kJ mol−1 is done as,
1 J mol−1= 10−3 kJ mol−1
Therefore, the conversion of 159.6×103 J mol−1 to kJ mol−1 is done as,
159.6×103 J mol−1=159.6 kJ mol-1_
(ii)
Interpretation: Under the given conditions, the energy of quantum has to be calculated.
Concept introduction: The quantum of energy is called ass photon. The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons. Photon is the fundamental particles of visible light. The energy of quantum is directly proportional to the frequency of radiation.
(ii)
Answer to Problem 7A.4BE
The energy of quantum in joule and kilo joule per mole are 3×10-19 J_ and 180.69 kJ mol-1_ respectively.
Explanation of Solution
The energy of quantum is directly proportional to the frequency of radiation
Therefore,
E=hν (1)
Where,
- E is the energy of quantum.
- ν is the frequency of radiation.
- h is the Planck’s constant (6.626×10−34 J s)
The frequency in terms of time period is expressed as,
ν=1T (2)
Where,
- T is the time period of electronic oscillation.
Substitute the value of ν in equation (1).
E=h×1T (3)
It is given that,
The time period is 2.21 fs.
The conversion of fs to s is done as,
1 fs=10−15 s
Therefore, the conversion of 2.21 fs to s is done as,
2.21 fs=2.21×10−15 s
Substitute the value of h and T in equation (3).
E=6.626×10−34 J s×12.21×10−15 s=3×10-19 J_
Per mole energy is calculated by using the formula,
ENA=NA×E (4)
Where,
- NA is Avogadro’s number (6.023×1023).
Substitute the value of NA and E in equation (4).
ENA=6.023×1023 mol−1×3×10−19 J =180.69×103 J mol−1
The conversion of J mol−1 to kJ mol−1 is done as,
1 J mol−1= 10−3 kJ mol−1
Therefore, the conversion of 180.69×103 J mol−1 to kJ mol−1 is done as,
180.69×103 J mol−1=180.69 kJ mol-1_
(iii)
Interpretation: Under the given conditions, the energy of quantum has to be calculated.
Concept introduction: The quantum of energy is called ass photon. The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons. Photon is the fundamental particles of visible light. The energy of quantum is directly proportional to the frequency of radiation.
(iii)
Answer to Problem 7A.4BE
The energy of quantum in joule and kilo joule per mole are 6.626×10-31 J_ and 39.9 ×10-11 kJ mol-1_ respectively.
Explanation of Solution
The energy of quantum is directly proportional to the frequency of radiation
Therefore,
E=hν (1)
Where,
- E is the energy of quantum.
- ν is the frequency of radiation.
- h is the Planck’s constant (6.626×10−34 J s)
The frequency in terms of time period is expressed as,
ν=1T (2)
Where,
- T is the time period of electronic oscillation.
Substitute the value of ν in equation (1).
E=h×1T (3)
It is given that,
The time period is 1.0 ms.
The conversion of ms to s is done as,
1 ms=10−3 s
Substitute the value of h and T in equation (3).
E=6.626×10−34 J s×110−3 s=6.626×10-31 J_
Per mole energy (ENA) is calculated by using the formula,
ENA=NA×E (4)
Where,
- NA is Avogadro’s number (6.023×1023).
Substitute the value of NA and E in equation (4).
ENA=6.023×1023 mol−1×6.626×10−31 J =39.9×10−8 J mol−1
The conversion of J mol−1 to kJ mol−1 is done as,
1 J mol−1= 10−3 kJ mol−1
Therefore, the conversion of 39.9×10−8 J mol−1 to kJ mol−1 is done as,
39.9×10−8 J mol−1=39.9 ×10-11 kJ mol-1_
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Chapter 7 Solutions
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
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