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wariber [46]
3 years ago
11

The hydrogen atom, changing from its first excited state to its lowest energy state, emits light with a wavelength of 122 nm. Th

at is in the far ultraviolet. The sodium atom, which like hydrogen has one electron that gets excited outside a core of 10 other electrons, emits light at 589 nm making a similar transition from its first excited state to its lowest state. Which of these statements would be true about the sodium and hydrogen atoms and their spectra?
a. There would be other series of spectral lines ending on the first excited states of both atoms.
b. There would be a series of spectral lines in hydrogen with the longest wavelength one at 122 nm.
c. There would be a series of spectral lines in sodium with the longest wavelength one at 589 nm.
d. The hydrogen atom binds its electron more tightly than the sodium atom does, and would require more energy to remove its electron completely.
Physics
1 answer:
tatyana61 [14]3 years ago
8 0

Answer:

true b and c

Explanation:

n the electromechanical transitions of the atoms the relationship must be fulfilled

        \frac{x}{\lambda } = R (1 / nf - 1 / no²)

where for the final state nf = 1 giving in the case of hydrogen the Lymma series whose smallest wavelength is lam = 122 nm with nf = 1 and there are a series of spectral lines for each value of n of the final state

in the case of sodium so well it has a transition from an excited state to the kiss state (bad)

Now let's review the different proposals

a) False. The electronic potential for sodium is much lower than for hydrognosia

b) True

c) True

d) true

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During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of
olasank [31]

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball m_A = 72.0 kg

mass of performer m_B = 65.0 kg

The horizontal component of the ball initially \mu_{xA} = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v

72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v

468 kg m/s + 0 = (137 kg)v

v = \dfrac{468\  kg m/s }{137 \ kg}

v = 3.416 m/s

8 0
3 years ago
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
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solmaris [256]

Answer:

Explanation:

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3 0
3 years ago
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Kobotan [32]

Answer:

6.14 s

Explanation:

The time the rocket takes to reach the top is only determined from its vertical motion.

The initial vertical velocity of the rocket is:

u_y = u sin \theta = (100)(sin 37^{\circ})=60.2 m/s

where

u = 100 m/s is the initial speed

\theta=37^{\circ} is the angle of launch

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v_y = u_y +gt

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v_y is the vertical velocity at time t

g=-9.8 m/s^2 is the acceleration of gravity

The rocket reaches the top when

v_y =0

So by substituting into the equation, we find the time t at which this happens:

t=-\frac{u_y}{g}=-\frac{60.2}{-9.8}=6.14 s

7 0
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Dominik [7]
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