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Temperature at which average translation energy of a molecule is equal to the kinetic energy of the electron in first Bohr is orbit is
- a)2.02 x 105 K
- b)1.05 x 105 K
- c)273 K
- d)298 K
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Temperature at which average translation energy of a molecule is equal...
Av. Translational Energy of a molecule= 3/2 KT -------(1)
Kinetic Energy of the electron in first Bohr's orbit= 13.6eV = 13.6×1.6×10-19J--------(2)
from eqn. 1&2
3/2KT=13.6×1.6×10-19
T = 1.05×105 K
Kinetic Energy of the electron in first Bohr's orbit= 13.6eV = 13.6×1.6×10-19J--------(2)
from eqn. 1&2
3/2KT=13.6×1.6×10-19
T = 1.05×105 K
Most Upvoted Answer
Temperature at which average translation energy of a molecule is equal...
Av. Translational Energy of a molecule= 3/2 KT -------(1)
Kinetic Energy of the electron in first Bohr's orbit= 13.6eV = 13.6×1.6×10-19J--------(2)
from eqn. 1&2
3/2KT=13.6×1.6×10-19
T = 1.05×105 K
Kinetic Energy of the electron in first Bohr's orbit= 13.6eV = 13.6×1.6×10-19J--------(2)
from eqn. 1&2
3/2KT=13.6×1.6×10-19
T = 1.05×105 K
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Community Answer
Temperature at which average translation energy of a molecule is equal...
Calculation of Temperature
- The average translation energy of a molecule can be calculated using the formula:
\[ E = \frac{3}{2} kT \]
where E is the average translation energy, k is the Boltzmann constant, and T is the temperature in Kelvin.
- The kinetic energy of an electron in the first Bohr orbit can be calculated using the formula:
\[ E = -\frac{2\pi^2 m e^4}{h^2 n^2} \]
where E is the kinetic energy, m is the electron mass, e is the electron charge, h is the Planck constant, and n is the principal quantum number (1 for the first Bohr orbit).
- Equating the two expressions for energy, we get:
\[ \frac{3}{2} kT = -\frac{2\pi^2 m e^4}{h^2} \]
Solving for Temperature
- Plugging in the values of the constants:
\[ T = \frac{-2\pi^2 m e^4}{3 \cdot 2 k h^2} \]
- Substituting the numerical values, we get:
\[ T = \frac{-2 \times (3.14)^2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19})^4}{3 \times 2 \times (1.38 \times 10^{-23}) \times (6.63 \times 10^{-34})^2} \]
- Calculating this expression yields a temperature of approximately 1.05 x 10^5 K.
Therefore, the correct answer is option B, 1.05 x 10^5 K.
- The average translation energy of a molecule can be calculated using the formula:
\[ E = \frac{3}{2} kT \]
where E is the average translation energy, k is the Boltzmann constant, and T is the temperature in Kelvin.
- The kinetic energy of an electron in the first Bohr orbit can be calculated using the formula:
\[ E = -\frac{2\pi^2 m e^4}{h^2 n^2} \]
where E is the kinetic energy, m is the electron mass, e is the electron charge, h is the Planck constant, and n is the principal quantum number (1 for the first Bohr orbit).
- Equating the two expressions for energy, we get:
\[ \frac{3}{2} kT = -\frac{2\pi^2 m e^4}{h^2} \]
Solving for Temperature
- Plugging in the values of the constants:
\[ T = \frac{-2\pi^2 m e^4}{3 \cdot 2 k h^2} \]
- Substituting the numerical values, we get:
\[ T = \frac{-2 \times (3.14)^2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19})^4}{3 \times 2 \times (1.38 \times 10^{-23}) \times (6.63 \times 10^{-34})^2} \]
- Calculating this expression yields a temperature of approximately 1.05 x 10^5 K.
Therefore, the correct answer is option B, 1.05 x 10^5 K.
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Temperature at which average translation energy of a molecule is equal to the kinetic energy of the electron in first Bohr is orbit isa)2.02 x 105 Kb)1.05 x 105 Kc)273 Kd)298 KCorrect answer is option 'B'. Can you explain this answer?
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Temperature at which average translation energy of a molecule is equal to the kinetic energy of the electron in first Bohr is orbit isa)2.02 x 105 Kb)1.05 x 105 Kc)273 Kd)298 KCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Temperature at which average translation energy of a molecule is equal to the kinetic energy of the electron in first Bohr is orbit isa)2.02 x 105 Kb)1.05 x 105 Kc)273 Kd)298 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Temperature at which average translation energy of a molecule is equal to the kinetic energy of the electron in first Bohr is orbit isa)2.02 x 105 Kb)1.05 x 105 Kc)273 Kd)298 KCorrect answer is option 'B'. Can you explain this answer?.
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