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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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