Question

If the velocity of an electron in Bohr's first orbit is $2.19\times {10}^{6}m{s}^{-1}$. What will be de Broglie wavelength associated with it?

• A. $2.19\times {10}^{-6}m$
• B. $4.38\times {10}^{-6}m$
• C. $3.32\times {10}^{-10}m$
• D. $3.32\times {10}^{10}m$

Answer 1

The correct option is C $3.32\times {10}^{-10}m$

According to de Broglie's equation

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

where, $\lambda =$ de-Broglie wavelength

$h=6.626\times {10}^{-34}Js$=Planck's constant

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

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

upon substitution we get:

$\lambda =\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 2.19\times {10}^6}$

$\lambda =0.332\times {10}^{-9}=3.32\times {10}^{-10}$m