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LiRa [457]
3 years ago
5

A hypothetical atom has three energy levels: the ground-state level and levels 1.50 eV and 5.00 eV above the ground state. What

is the longest wavelength in the line spectrum for this atom? Let Planck's constant h = 4.136 x 10^-15 ev s, and the speed of light c = 3.00 x 10^8 m/s. (a)355 nm (b)780 nm (c)882 nm (d) 827 nm
Physics
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

option (d)

Explanation:

E1 = 1.5 eV = 1.5 x 1.6 x 10^-19 J, E2 = 5 eV = 5 x 1.6 x 10^-19 J, c = 3 x 10^8 m/s, h = 6.62 x 10^-34 Js

Wavelength associated with 1.5 eV is λ1.

E1 = h c / λ1

λ1 = h c / E1

λ1 = (6.62 x 10^-34 x 3 x 10^8) / (1.5 x 1.6 x 10^-19)

λ1 = 8.275 x 10^-7 m = 827 nm

Wavelength associated with 5 eV is λ2.

E2 = h c / λ2

λ2 = h c / E2

λ2 = (6.62 x 10^-34 x 3 x 10^8) / (5 x 1.6 x 10^-19)

λ2 = 2.4825 x 10^-7 m = 248 nm

So, the longest wavelength is 827 nm

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Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 3.
noname [10]

Explanation:

The speed of a wave is given by :

v=f\lambda ......(1)

(i) Here, the wavelength of a sound wave in air reduces by a factor of 3. Equation (1) becomes :

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Which two particles are present in the nucleus of an atom? Electrons and neutrons Electrons and molecules 1 TH O Protons and neu
aleksandrvk [35]

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8 0
3 years ago
An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

So we have v_{AT}=(95km/h)-(75Km/h)=20Km/h.

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s

And in that time the car would have traveled (<em>relative to the ground</em>):

d=v_At=(95Km/h)(0.065h)=6.175Km

If they are traveling in opposite directions, <u>we have to do all the same</u> but using v_{AT}=v_A+v_T (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

v_{AT}=(95km/h)+(75Km/h)=170Km/h

t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

5 0
3 years ago
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