Ask question

Ask question

All categories

2 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?

1 answer:

Troyanec [42]2 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:

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

You might be interested in

A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using $\pi r^{2} \\$ where r is the radius of 5cm or 0.05m

$A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\$

since we no that the angle is at $0^{0}$

we determine the magnitude of the magnetic filed

$B=0.5e^{-t} \\t=2s$

$E=-(0.5e^{-2} * 0.00785)$

$E=-0.000532v\\$

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

$V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A$

$I=2.6mA$

6 0

3 years ago

The "atomic weight" of an atom reflects the average number of

Illusion [34]

E) Protons, neutrons, and electrons

8 0

2 years ago

The field of a solenoid is most like the field of a: long, straight wire single bar magnet horseshoe magnet single pole

user100 [1]

The outside magnetic field of a solenoid is similar to that of a bar magnet even though the inside field is strong and uniform.

4 0

2 years ago

Read 2 more answers

An electron of mass 9.11×10-31kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to t

SOVA2 [1]

Answer:

(A) Acceleration will be $240.3846\times 10^{12}m/sec^2$

(b) Time taken will be $1.4\times 10^{-8}sec$

(c) Force will be $2189.9\times 10^{-19}N$

Explanation:

We have given that electron starts from rest so initial velocity u = 0 m/sec

Final velocity $v=2.50\times 10^6m/sec$

Mass of electron $m=9.11\times 10^{-31}kg$

Distance traveled by electron $s=1.30cm =0.013m$

From third equation of motion we know that $v^2=u^2+2as$

(a) So $(2.5\times 10^6)^2=0^2+2\times a\times 0.013$

$a=240.3846\times 10^{12}m/sec^2$

(b) From first equation of motion we know that v = u+at

So $2.50\times 10^6=0+240.3846\times 10^{12}t$

$t=0.014\times 10^{-6}=1.4\times 10^{-8}sec$

(c) From newton's law we know that force

$F=ma=9.11\times 10^{-31}\times 240.3846\times 10^{12}=2189.9\times 10^{-19}N$

7 0

2 years ago

An object moves in one dimension according to the function x(t)=13at3, where a is a positive constant with units of ms3. During

9966 [12]

Answer:

B) 1/5 ba^2 T^5

Explanation:

The dissipated energy is given by the work done over the object by the force F=-bv. The work is given by the following formula:

$dW=Fdx$

you derivative the function f(x) and replace v by the derivative dx/dt you obtain:

$v=\frac{dx}{dt}=at^2\\\\dx=at^2dt\\\\W=\int_0^{T} Fdx=-\int_0^Tvbdx=-\int_0^Tb(at^2)(at^2dt)\\\\W=-ba^2\frac{T^5}{5}=-\frac{1}{5}ba^2T^5$

hence, the dissipated energy is 1/5 ba^2 T^5

7 0

3 years ago