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slava [35]
3 years ago
12

The spectrum of a distant star shows that one in 2 e6 of the atoms of a particular element is in its first excited state 7.5 eV

above the ground state. What is the temperature of the star? (You can ignore the other excited states and assume the ratio of statistical weights is 4
Physics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

The temperature of star is 5473.87 K

Explanation:

Given:

Energy difference \Delta E = 7.5 eV

The ratio of number of particle \frac{N_{f} }{N_{i} } = \frac{1}{2 \times 10^{6} }

Degeneracy ratio \frac{g_{f} }{g_{i} }  = 4

From the formula of boltzmann distribution for population levels,

     \frac{N_{f} }{N_{i} } =\frac{g_{f} }{g_{i} }  e^{-\frac{\Delta E}{kT} }

Where k = boltzmann constant = 8.62 \times 10^{-5} \frac{eV}{K}

     \frac{1}{2 \times 10^{6} } =4  e^{-\frac{7.5 eV}{8.62 \times 10^{-5} T} }

  8 \times 10^{6} } = e^{\frac{7.5 eV}{8.62 \times 10^{-5} T} }

\ln(8 \times 10^{6})  = {\frac{7.5 eV}{8.62 \times 10^{-5} T} }

  T  = {\frac{7.5 eV}{8.62 \times 10^{-5} \ln(8 \times 10^{6})} }

  T = 5473.87 K

Therefore, the temperature of star is 5473.87 K

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Answer:

\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

Explanation:

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q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

where the cross product can be made with the determinant method.

Hope this helps!!

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