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harkovskaia [24]
3 years ago
9

When an excited electron in a hydrogen atom falls from the n = 3 to the n = 2 level, a photon of red light is emitted from the a

tom. Compared to the red photon, what would be true about the electromagnetic radiation emitted if an electron were to fall from n = 4 to n = 3?

Physics
1 answer:
Kaylis [27]3 years ago
3 0

Answer: 653.33 nm ; 1875, 24 nm

Explanation: For the first case we have to use the Balmer series for  the hydrogen  when the atom falls from the n = 3 to the n = 2. So for the second transtions for the hydrogen we use  the Paschen serie.   To do  the calculation we need to know  the Ryberg constant that is equal to 1.097 * 10^7 m^-1. In the attach is shown the expression for spectral series used for calculation.

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A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters
Tanya [424]

Answer:

The change in potential energy of the mass as it goes up the incline is 0.343 joules.

Explanation:

We must remember in this case that change in the potential energy is entirely represented by the change in the gravitational potential energy. From Work-Energy Theorem and definition of work we get that:

U_{g}= m\cdot g\cdot \Delta y

Where:

U_{g} - Gravitational potential energy, measured in Joules.

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

\Delta y - Change in vertical height, measured in meters.

This work is the energy needed to counteract effects of gravity at given vertical displacement.

If we know that m = 0.5\,kg, g = 9.807\,\frac{m}{s^{2}} and \Delta y = 0.07\,m, the change in the potential energy of the mass as it goes up the incline is:

U_{g} = (0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.07\,m)

U_{g} = 0.343\,J

The change in potential energy of the mass as it goes up the incline is 0.343 joules.

5 0
3 years ago
What will happen to the 0.1 N force if one charges is increased by a factor of 3?
AleksandrR [38]

Answer:

Explanation:

Force between two charges is given by the following expression

F = \frac{KQ_1Q_2}{d^2}  Q₁ and Q₂ are two charges and d is distance between two.

.1 = \frac{KQ_1Q_2}{d^2}

If Q₁ becomes three times , force will become 3 times . Hence force becomes .3 N in the first case.

Force F = .3 N

If charge becomes one fourth , force also becomes one fourth .

F= \frac{.1}{4}

= .025 N.

5 0
3 years ago
The policeman also watched a truck that was sitting at the red light. When the light turned green, the truck slowly started movi
DerKrebs [107]

Answer:

a steady level climb until he gets to his constant speed

Explanation:

a steady level climb until he gets to his constant speed

3 0
4 years ago
Read 2 more answers
what must be the distance between source of sound and mountain to give an echo in 5 seconds ? ,(speed of sound at 0°c=33m\sec​
Tamiku [17]

Answer:

distance must be = 330 × 5/2

= 330×2.5

=825m

4 0
3 years ago
42,43 and 44 please
Mariana [72]
42) The sailboat travels east with velocity v_e=30 mph, while the current moves south with speed v_s=30 mph. Since the two velocities are perpendicular to each other, he resultant velocity will be given by the Pytagorean theorem:
v= \sqrt{v_e^2+v_s^2}= \sqrt{(30)^2+(30)^2}=42 mph
and the direction is in between the two original directions, therefore south-east. So, the correct answer is
D) 42 mph southeast

43) Since the light moves by uniform motion, we can calculate the distance corresponding to one light year by using the basic relationship between velocity, space and time. In fact, we know the velocity:
v=300,000,000 m/s=3 \cdot 10^8 m/s
and the time is one year, corresponding to:
t=32,000,000 s=3.2 \cdot 10^7 s
therefore, the distance corresponding to one light year is:
S=vt=(3\cdot 10^8 m/s)(3.2 \cdot 10^7 s)=9.6 \cdot 10^{15}m
Therefore, the correct answer is D.

44) For the purpose of the problem, we can assume that the light travels instantaneously from the flash to us (because the distances involved are very small), so the time between the flash and the thunder corresponds to the time it took for the sound to travel to us.
The speed of sound is
v=1100 ft/s=750 mph=335 m/s
And since the time between the flash and the thunder is t=3 s, the distance is
S=vt=(335 m/s)(3 s)=1005 m=0.62 miles
Therefore, the correct answer is A) 3/5 mile.
6 0
3 years ago
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