Question

If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6} ms^{-1}$, calculate the de Broglie wavelength associated with it.

Answer 1

"According to De-Broglie's equation,  

$\lambda \quad =\quad \frac { h }{ mv } \quad ......(i)$

Here, $h\quad =\quad 6.626\quad \times \quad { 10 }^{ -34 }\quad kg.{ m }^{ 2 }{ s }^{ -1 };$

$v\quad =\quad 2.19\quad \times \quad { 10 }^{ 6 }\quad m{ s }^{ -1 };$

$m\quad =\quad 9.11\quad \times \quad { 10 }^{ -31 }\quad kg$

Substituting the values in the equation (i), we get

$\lambda \quad =\quad \frac { 6.626\quad \times \quad { 10 }^{ -34 }\quad kg{ m }^{ 2 }{ s }^{ -1 } }{ \left( 9.11\quad \times \quad { 10 }^{ -31 }\quad kg \right) \quad \times \quad \left( 2.19\quad \times \quad { 10 }^{ 6 }\quad m/s \right)}$

$\lambda \quad =\quad 3.32\quad \times \quad { 10 }^{ -10 }\quad m\quad =\quad 332\quad pm$"

Answer 2

Answer:

According to de Broglie's equation

λ=

mv

h

where, λ= de-Broglie wavelength

h=6.626×10

−34

Js=Planck's constant

m=mass of electron=9.1×10

−31

kg

velocity of electron, v=2.19×10

6

m/s

upon substitution we get:

λ=

9.1×10

−31

×2.19×10

6

6.626×10

−34

λ=0.332×10

−9

=3.32×10

−10

m