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drek231 [11]
3 years ago
13

As mentioned in this week’s notes on page 4, the electrons of an atom can occupy different energy shells within the atom (simila

r to how the planets all occupy different orbits around the Sun). Electrons prefer to be in the lowest energy shell possible (the ground state); however, they can gain energy and jump to a higher shell by absorbing light or being excited by an electric current. In accordance with the conservation of energy, if an electron drops from a higher energy level to a lower one, this must emit a photon (particle of light) with energy equal to the energy difference of the shells.
A Balmer series transition is any transition of an electron from some higher energy shell down to the second lowest energy shell (n=2) in hydrogen.

Looking at image (b) above, what is the wavelength of a photon emitted during the Balmer transition from the n=3 shell in hydrogen? (remember nm is short for a nanometer, for example 656 nm = 656 x 10-9 meters)

A) 656E-9 meters
B) 486E-9 meters
C) 434E-9 meters
D) 410E-9 meters

Physics
1 answer:
Nostrana [21]3 years ago
7 0
The assume that goes with the inquiry demonstrates the wavelenghts of the photons transmitted by Balmer arrangement change , from vitality levels (n) 3, 4, 5, and 6 to the vitality level (n) 2, in  hydrogen particles.

These are the values shown in the figure

Transition           wavelength of the photon emitted
                                   nm


from n=3  to n=2              656<------ this is the value requested

from n=4 to n=2               486

from n=5 to n=2               434

from n=6 to n=2              410

The wavelength of a photon discharged from the n = 3 shell in hydrogen is the primary information of the table, i.e 656 nm.

Using the conversion factor from nm to m that results is :

656 nm * 1 m / (10^9 nm) = 656 * 10 ^ -9 m.
Hope this helps
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An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t
lana [24]

Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

q_{h} = R* T_{h} in(V_{2}/V_{1})

   =0.287 * 1200 ln(3)

   =0.287*1318.33

   =378.36kJ/kg

The specific heat capacity:

q_{L}=q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

n_{TH}=1 - (T_{L}/T_{H})

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

4 0
3 years ago
When an object's final velocity is less than its initial velocity, however, it has ________________ acceleration.
algol [13]

Answer:

The body has negative acceleration PR a deceleration.

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0
3 years ago
A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet
Vikki [24]

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

3 0
3 years ago
A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

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f-N_w = 0

N_w = f

The forces in the horizontal direction would be,

\sum F_y = 0

N_f -W =0

N_f = W

The sum of Torques at equilibrium,

\sum \tau = 0

Wdcos\theta - N_wLsin\theta = 0

WdCos\theta = fLSin\theta

f = \frac{Wd}{Ltan\theta}

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

f_{max} = \mu W=\frac{Wd}{Ltan\theta}

\theta = tan^{-1} (\frac{d}{\mu L})

Replacing,

\theta = tan^{-1} (\frac{0.9}{0.42*2})

\theta = 46.975\°

Therefore the minimum angle that the person can reach is 46.9°

8 0
3 years ago
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