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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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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
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(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

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(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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If an airplane were traveling eastward with a thrust force 450 N and there was a tailwind of 200 N, what would the resulting net
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The resulting net force is 650N

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Drag and tail force on a plane :brainly.com/question/11922080

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