Question

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display the equality in your ranking. (a) a photon with energy 3eV(b) an electron with kinetic energy 3eV(c) a proton with kinetic energy 3eV(d) a photon with energy 0.3eV(e) an electron with momentum 3eV/c

Answer 1

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

• De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
• From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

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) .

Learn more about de brogile here :

brainly.com/question/28165547

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