When smoke particles pass between the source of radiation and the detector the drop in radiation will be sufficient to tigger the alarm. <span />
Answer:
o to increase the frequency of sound waves. It increases the sound waves to a level of frequency that humans cannot hear so you won't be able to hear many things though the wall other then low noises like pounding.
Explanation:
I am in construction class as well as a student teacher for other construction type programs trust me :D
Brainiest would be appreciated
Answer:
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Explanation: The law of copying
Answer:
The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Explanation:
Given;
wavelength of ultraviolet light, λ = 270 nm
work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J
The energy of the ultraviolet light is given by;
![E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B%286.626%2A10%5E%7B-34%7D%20%29%283%2A10%5E%7B8%7D%29%20%7D%7B270%2A10%5E%7B-9%7D%20%7D%5C%5C%5C%5CE%20%3D%207.362%20%2A%2010%5E%7B-19%7D%20%5C%20J)
The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;
E = φ + K.E
K.E = E - φ
K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )
K.E = 3.677 x 10⁻¹⁹ J
K.E = ¹/₂mv²
mv² = 2K.E
velocity of the electron is given by;
![V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s](https://tex.z-dn.net/?f=V%20%3D%20%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%20%7D%5C%5C%5C%5CV%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%283.677%2A10%5E%7B-19%7D%29%20%7D%7B9.1%2A10%5E%7B-31%7D%20%7D%20%7D%5C%5C%5C%5CV%20%3D%208.99%2A10%5E%7B5%7D%20%20%5C%20m%2Fs)
the shortest de Broglie wavelength for the electrons is given by;
![\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bh%7D%7Bmv%7D%5C%5C%20%5C%5C%5Clambda%20%3D%20%5Cfrac%7B6.626%2A10%5E%7B-34%7D%20%7D%7B%289.1%2A10%5E%7B-31%7D%29%28%208.99%2A10%5E%7B5%7D%20%29%7D%5C%5C%5C%5C%5Clambda%20%3D%208.10%2A10%5E%7B-10%7D%20%5C%20m%5C%5C%5C%5C%5Clambda%20%3D%200.81%20%5C%20nm)
Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Answer:Theoretical Discussion
The diffraction of classical waves refers to the phenomenon wherein the waves encounter an obstacle that fragments the wave into components that interfere with one another. Interference simply means that the wave fronts add together to make a new wave which can be significantly different than the original wave. For example, a pair of sine waves having the same amplitude, but being 180◦ out of phase will sum to zero, since everywhere one is positive, the other is negative by an equal amount.