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Triss [41]
2 years ago
12

In a ruby laser, an electron jumps from a higher energy level to a lower one. if the energy difference between the two levels is

1.8 ev, what is the wavelength of the emitted photon?
Physics
1 answer:
Simora [160]2 years ago
4 0

In a ruby laser, an electron jumps from a higher energy level to a lower one. if the energy difference between the two levels is 1.8 EV, 649 is the wavelength of the emitted photon

Only pulse mode is employed with the ruby laser. Water does not readily absorb radiation, but pigments like melanin and hemoglobin do so to a large extent. The anterior ocular structures are easily penetrated by the ruby laser. Vascular and colored retinal lesions are photocoagulation with it.

Ruby served as the active medium in Maiman's 1960 development of the solid-state laser known as the Ruby laser. Aluminum oxide crystals are found in rubies. where chromium ions are used to partially replace aluminum ions. Chromium ions are the substance that makes the ruby laser active.

To learn more about ruby laser please visit
brainly.com/question/28168316
#SPJ4

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Have a positive effect on you as a human being. Your attitude will improve, your charisma will spread, and you will a better, more well rounded person.
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4 years ago
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Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 494 Hz when air temperature i
elena-14-01-66 [18.8K]

Answer:

0.173 m.

Explanation:

The fundamental frequency of a closed pipe is given as

fc = v/4l .................. Equation 1

Where fc = fundamental frequency of a closed pipe, v = speed of sound  l = length of the pipe.

Making l the subject of the equation,

l = v/4fc ................ Equation 2

also

v = 331.5×0.6T ................. Equation 3

Where T = temperature in °C, T = 18.0 °c

Substitute into equation 3

v = 331.5+0.6(18)

v = 331.5+10.8

v = 342.3 m/s.

Also given: fc = 494 Hz,

Substitute into equation 2

l = 342.3/(4×494)

l = 342.3/1976

l =0.173 m.

Hence the length of the organ pipe = 0.173 m.

7 0
3 years ago
Please help me to solve this and give me the summary of answer
Neko [114]
Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

                     u = w

                 vdg =  mg      Cancel out g

                   vd  =  m
 
The v  is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

         0.2V * 1000 =  m

           200V =  m

Hence the mass of the object is  200V  kg.

But Density of solid =  Mass of solid / Volume of solid

                                 =    200V / V

                                 =    200 kg/m³

Density of solid = 200 kg/m³     
7 0
3 years ago
A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance
Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, m_b=11\ g=0.011\ kg

Mass of the pendulum, m_p=19\ kg

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)

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Put the value of V from equation (1) in above equation as :

v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s

So, the bullet's initial speed is 243.21 m/s.

7 0
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