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Lapatulllka [165]
3 years ago
5

In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t

he Balmer series). If hydrogen absorption lines are very strong in the visible spectrum of a particular star, what is this star's surface temperature?
1. relatively low temperature, about 6000 K (similar to the Sun), because at lower temperatures too many hydrogen atoms are in the ground (n = 1) state, and at higher temperatures too many have had their electrons completely ejected.
2. very high temperature, about 40,000 K, because at lower temperatures too many hydrogen atoms have their electrons in the ground state (n = 1) and cannot absorb in the Balmer series.
3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms.
4. very low temperature, 3000 K, because at higher temperatures so many hydrogen atoms have had their electrons completely ejected and cannot absorb in the Balmer series.
Physics
1 answer:
grigory [225]3 years ago
4 0

Answer:

3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms

Explanation:

If hydrogen absorption lines are very strong in the visible spectrum of a particular star that means the population of electron in n = 2 is very high so on being exited they absorb radiation in Balmer series and give rise to absorption spectrum. The average temperature required to excite electron in hydrogen atom from n=1 to n = 2   is 10000K .

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<h3>Further explanation</h3>

Given

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SOVA2 [1]

Answer:

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Explanation:

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So,  Formula for energy:

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Read 2 more answers
A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

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we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

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3 years ago
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