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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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Answer:

N = 195 turns

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The length of the tube, l = 12 cm = 0.12 m

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\mu_{0} = 4\pi * 10^{-7}

The inductance of a solenoid is given by:

L = \frac{\mu_{0}N^{2} A }{l}

500 * 10^{-6} = \frac{4\pi *10^{-7}  N^{2} *4\pi  *10^{-4}  }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns

8 0
2 years ago
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

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         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

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let's reduce to revolutions / s

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Answer:

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A 30kg mass is brought from the earth to the moon what is the weight on the earth
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Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

2

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hodyreva [135]

i guess option (a)

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