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Novay_Z [31]
2 years ago
14

Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with

Z=5 and only a single electron orbiting the nucleus. This expression predicts equal electron energies for these two species for certain values of the quantum number n (the quantum number is different for each species). For quantum numbers less than or equal to 9, what is the highest energy (in electron volts) for which the helium energy level is equal to the ionized atom energy level?
Physics
1 answer:
ella [17]2 years ago
3 0

Answer:

Explanation:

Bohr's energy expression is as follows

E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .

z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom

so

13.6 x 2² / n₁² = 13.6 x 5² / n₂²

n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .

For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He

Putting it in the equation above

2² / 8² = 5² / n₂²

n₂ = 5 x 8 / 2

= 20 .

energy

= -  13.6 x2² / 8²

= -  0.85 eV .

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Complete combustion of 1.0 metric ton of coal (assuming pure carbon) to gaseous carbon dioxide releases 3.3 x 1010 j of heat. co
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6 0
2 years ago
In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?
faust18 [17]
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A. diffraction 
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<span>C. reflection </span>
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8 0
2 years ago
A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106
defon

Answer:

4.7\cdot 10^{-14}N

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

F=qvB

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

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B=4.5\cdot 10^{-2} T

So the force experienced by the electrons is

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3 0
2 years ago
You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7°  northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

V_{y} = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

V_{x} = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{13.89^{2} +9.89^{2}   } = <em>17.05 mph  This is your airspeed</em>

b) To get your direction, we use

tan ∅ = V_{y} /V_{x}

tan ∅ = 13.89/9.89 = 1.413

∅ = tan^{-1}(1.413) = <em>54.7°  northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

V_{y} = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{4.11^{2} +9.89^{2}   } = <em>10.71 mph  This is your airspeed</em>

Your direction will be,

tan ∅ = V_{y} /V_{x}

tan ∅ = -4.11/9.89 = -0.416

∅ = tan^{-1}(-0.416) =<em> -22.58°  this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

5 0
2 years ago
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