Question
The de-Broglie wavelength of a particle with mass $$1 g$$ and velocity $$100 m/s$$ is
A.
$$6.63 \times {10^{ - 33}}m$$
B.
$$6.63 \times {10^{ - 34}}m$$
C.
$$6.63 \times {10^{ - 35}}m$$
D.
$$6.65 \times {10^{ - 36}}m$$
Answer :
$$6.63 \times {10^{ - 33}}m$$
Solution :
$$\eqalign{
& p = \frac{h}{\lambda }{\text{(de - Broglie equation)}} \cr
& \lambda = \frac{h}{{mv}}\,\,\left( {\because \,\,p = mv} \right) \cr
& h = 6.625 \times {10^{ - 34}} \cr
& \,\,\,\,\, \approx 6.63 \times {10^{ - 34}}kg/s \cr
& \lambda = \frac{{6.63 \times {{10}^{ - 34}}kg\,{m^2}/s}}{{{{10}^{ - 3}}kg \times 100m/s}} \cr
& \,\,\,\,\, = 6.63 \times {10^{ - 33}}m \cr} $$