Question
If the velocity of an electron in Bohr's first orbit is $$2.19 \times {10^6}\,m\,{s^{ - 1}},$$ what will be the 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 :
$$3.32 \times {10^{ - 10}}\,m$$
Solution :
$$\eqalign{
& \lambda = \frac{h}{{mv}} \cr
& \,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{9.11 \times {{10}^{ - 31}}\,kg \times 2.19 \times {{10}^6}\,m\,{s^{ - 1}}}} \cr
& \,\,\,\,\, = 3.32 \times {10^{ - 10}}\,m \cr} $$