Question

The radius of the second orbit of an electron in hydrogen atom is $2.116\stackrel{o}{A}$. The de Broglie wavelength associated with this electron in this orbit would be

• A. $6.64\stackrel{o}{A}$
• B. $1.058\stackrel{o}{A}$
• C. $2.116\stackrel{o}{A}$
• D. $13.28\stackrel{o}{A}$

Answer 1

The correct option is A $6.64\stackrel{o}{A}$

From Bohr's postulate, $mvr=\frac{nh}{2\pi }$ ... (1) where $m=$ mass of electron, $v=$ orbital velocity of electrons , $r=$ radius of the orbit and $h=$ Planck's constant

de Broglie wavelength, $\lambda =\frac{h}{p}=\frac{h}{mv}$ ....(2)

From (1) and (2), $\lambda =\frac{2\pi r}{n}=\frac{2\times 3.14\times 2.116}{2}=6.64A^o$