Question

The De-Broglie wavelength associated with electrons revolving round the nucleus in a hydrogen atom in ground state, will be-

• A. $0.3A^{\circ }$
• B. $3.3A^{\circ }$
• C. $6.62A^{\circ }$
• D. $10A^{\circ }$

Answer 1

The correct option is B $3.3A^{\circ }$

According to $Boh{r}^{\prime }s$ postulate $mvr=\frac{nh}{2\pi }$ so the momentum in ground state$(n=1)$ will be $p=mv=\frac{h}{2\pi r}$

Required wavelength will be $\lambda =\frac{h}{p}=\frac{h}{h/2\pi r}=2\pi r$

Value of radius for ground state in hydrogen is $r=0.53Angstrom$

Using that we get $\lambda =3.32Angstrom$

Option B is correct.