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MatroZZZ [7]
3 years ago
6

If we have a one-electron atom, and the electron can occupy any of 5 different energy levels, and transitions are possible betwe

en any level and any other, how many lines might appear in that atom’s spectrum?
Physics
1 answer:
xeze [42]3 years ago
7 0

Answer:

10

Explanation:

 Lets say that a,b,c,d,e are the five allowed energy states in order of decreasing energy. Then the number of possible different spectral lines comes from the electron dropping from a high state to a lower state. So, they will do so in following ways:

a - b

a - c

a - d

a- e

b - c

b - d

b- e

c - d

c- e

d- e

Ans- Ten different possible energy jumps giving six different colors or lines in the spectrum.

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O.25 m is the displacement
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3 years ago
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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

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3 years ago
An iron ball is dropped at a height of 10 m from the surface of the moon.
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Answer:

3.51s

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A strong inductive argument must have true premises<br><br> True<br><br> False
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That is true imo not false
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What is the speed of a truck traveling 10km in 10 minutes
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Answer: 58.8235 km/h

speed = distance/time

the distance is 10 km

the time is 10 minutes

the unit is not correct, so we first change minute to hour

so 10/60 is 0.166667, rounded to 0.17.

10 km/ 0.17 hours =

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