Question

The debroglie wavelength of an electron moving with velocity v = 3c/5 is

Answer 1

Answer:

The de Broglie wavelength of an electron moving with velocity v = 3c/5 is $0.40*10^{-11}m$

Explanation:

• The de Broglie wavelength is the wavelength of matter in wave nature

• For a particle with mass (m) and moving with velocity (v), the de Broglie wavelength is given by the formula

       λ =$h/p=h/mv$(equation 1)

      where Planck's constant $h=6.6*10^{-34}$

     $p$- the momentum of the particle

From the question, the given particle is an electron

the velocity of the electron $v=3C/5 m/s$

mass of the electron $m=9.1*10^{-31}kg$

put all the values in equation 1

⇒λ= $\frac{6.6*10^{-34} }{9.1*10^{-31}*3c/5 }=\frac{33}{81.9}*10^{-11} =0.40*10^{-11} m$

where the velocity of light $C=3*10^{8}m/s$

Answer 2

Answer:

c=3x10^8 m/s

Explanation:

De Broglie wavelength is an vital idea whilst reading quantum mechanics. The wavelength (λ) this is related to an item on the subject of its momentum and mass is referred to as de Broglie wavelength. A particle's de Broglie wavelength is commonly inversely proportional to its force.

The deBroglie wavelength is described as follows: lambda = h/mv , wherein the greek letter lambda represents the wavelength, h is Planck's contant, m is the particle's mass and v is its velocity.

The de Broglie wavelength is the wavelength of rely in wave nature

For a particle with mass (m) and shifting with velocity (v), the de Broglie wavelength is given through the formula

 λ =h/p

 λ =h/mv

wherein Planck's regular h=6.6x10^-34

p-the momentum of the particle

From the question, the given particle is an electron

the rate of the electron v=3c/5

mass of the electron=9.1 x 10 ^-34

 λ=6.6 x 10 ^-34/9.1 x 10 ^-34x3c/5

λ=0.40x10^-11

c=3x10^8m/s

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