Question

If the velocity of the electron in Bohr’s first orbit is 2.19 × 10^6 ms^–1, calculate the de Broglie wavelength associated with it.
Ncert solutions for Class 11th Chemistry Part - 1 Chapter 2 Exercise 59

Answer 1

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

Answer 2

Answer:

The de Broglie wavelength associated with it is 3.311 angstrom.Explanation:

De-Broglie wavelength is calculated by using the formula:

$\lambda=\frac{h}{mv}$

where,

$\lambda$ = wavelength of electron

h = Planck's constant = $6.6\times 10^{-34}Js$

m = mass of electron = $9.1\times 10^{-31}kg$

v = velocity of electron = $2.19\times 10^6 m/s$

$\lambda =\frac{6.6\times 10^{-34}Js}{9.1\times 10^{-31}kg\times 2.19\times 10^6 m/s}=.03311\times 10^{-9} m=3.311 \AA$

The de Broglie wavelength associated with it is 3.311 angstrom.