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lakkis [162]
3 years ago
5

A sodium atom will absorb light with a wavelength near 589 nm if the light is within 10 MHz of the resonant frequency. The atomi

c mass of sodium is 23. (i) Calculate the number of "yellow" photons of wavelength 2 = 589 nm that must be absorbed to stop a sodium atom initially at room temperature (V-600 m/s). [7 marks] (ii) What is the minimum time needed to cool a sodium atom?
Physics
1 answer:
Troyanec [42]3 years ago
7 0

Answer:

i)20369 photons

ii) 40 ps

Explanation:

Momentum of one Sodium atom:

P=m*v =600m/s*23amu*\frac{1 kg}{6.02*10^{23}amu}\\P=2.29*10^{-23}kgm/s

In other to stop it, it must absorb the same momentum in photons:

P=2.29*10^{-23}kgm/s=n_{photons}*\frac{h_{planck}}{\lambda}\\=n*\frac{6.63*10^{-34}}{589*10^{-9}} \\==>n=20369 photons

Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:

t=n*T=n*\frac{\lambda}{c} =20369*\frac{589nm}{3*10^{8}m/s}=4*10^{-11} second=40 ps

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3 years ago
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

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frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
3 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
What factors determine the amount of air resistance an object experiences?
geniusboy [140]
The pressure of the air at the way its blowing

3 0
3 years ago
A note of frequency 2000Hz has a velocity of 400 ms! What is the wavelength of the note?​
Rudik [331]

Answer:

Explanation:

frequency=c/λ

c=400m/s

putting values

2000=400/λ

λ=0.2m

8 0
3 years ago
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