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Travka [436]
3 years ago
12

hey can anybody tell me why my eyes water and burn when I go to stores I thoight that it was the cold air conditioner in stores

that caused that but today I msde my house really cold using ac and my eyes didnt burn or water as much so im thinking might it be an allergy to the air conditioner in the stores???? im really confused
Physics
1 answer:
bekas [8.4K]3 years ago
4 0
Its most likely a combination of the bright lights and the air vents. It may also be just because you get tired or drowsy in stores.
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The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

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De Broglie's relationship is given by  \lambda=\frac{h}{mv} .

This can be written as \lambda=\frac{h}{p}  .....(1) where λ  is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by p=\frac{E}{c}     ...(2)  where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

              p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m

Putting this value of p in equation (1) , we get

  \lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}

(b) As we know that  kinetic energy is given by

         K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p      ...(3)

Where mass of electron is 9.1\times10^-^3^1 kg .

Putting K.E = 3eV in equation (3) , we get

    \sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}

(c) Putting m=1.67\times10^{-27}kg and K.E = 3eV in equation (3) , we get

     \sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}

(d) Putting E = 0.3eV in equation (2) , we get

   p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}

(e) Putting p=3eV/c in equation (1) , we get

    \lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

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