Question

If the velocity of an electron in Bohr's first orbit is 2.19×106ms−1. What will be de Broglie wavelength associated with it?

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Answer 1

Answer:

v = 2.19 * 10⁶  m/s

λ = h / p

h = Planck's constant = 6.636 * 10⁻³⁴  units

m = mass of an electron = 9.11 * 10⁻³¹  kg

Bohr's first orbit  :  n = 1,  K shell in an atom... like Hydrogen atom..

λ = 6.636 * 10⁻³⁴ / [ 9.11 * 10⁻³¹ * 2.19 * 10⁶ ]    meters

  =  0.3326 * 10⁻⁹ meters

hope it will help u

Answer 2

Answer :

• de Brogile wavelength = 332 pm.

Step-by-step explanation :

Given that,

• Velocity of electron, v = $\rm{2.19\times{}10^6\:ms^{-1}}$

Also,

• Mass of electron, m = $\rm{9.11\times{}10^{-31}\:kg}$
• Planck's constant, h = $\rm{6.626\times{}10^{-34}\:kgm^2s^{-1}}$

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We know, according to de Broglie's equation,

$\rm{\lambda=\dfrac{h}{mv}}$

Putting the given values, we get,

$\implies\rm{\lambda=\dfrac{h}{mv}=\dfrac{6.626\times{}10^{-34}\:kgm^2s^{-1}}{(9.11\times{}10^{-31}\:kg)(2.19\times{}10^6\:ms^{-1})}}$

$\implies\rm{\lambda=3.32\times{}10^{-10}\:m=}\:\bold{332\:pm.}$