Question
The momentum of a particle having a de-Broglie wavelength of $${10^{ - 17}}m$$ is
$$\left( {{\text{Given,}}\,h = 6.625 \times {{10}^{ - 34}}m} \right)$$
A.
$$3.3125 \times {10^{ - 7}}kg\,m\,{s^{ - 1}}$$
B.
$$26.5 \times {10^{ - 7}}kg\,m\,{s^{ - 1}}$$
C.
$$6.625 \times {10^{ - 17}}kg\,m\,{s^{ - 1}}$$
D.
$$13.25 \times {10^{ - 17}}kg\,m\,{s^{ - 1}}$$
Answer :
$$6.625 \times {10^{ - 17}}kg\,m\,{s^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{According to de - Broglie relation,}} \cr
& \lambda = \frac{h}{{mv}} = \frac{h}{p} \cr
& {\text{where,}}\,\,\lambda = {\text{wavelength}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,h = {\text{Planck's}}\,{\text{constant}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p = {\text{momentum}} \cr
& {\text{Here,}}\,\,h = 6.625 \times {10^{ - 34}}J\,s \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda = {10^{ - 17}}m \cr
& \therefore \,\,p = \frac{h}{\lambda } \cr
& \,\,\,\,\,\,\,\,\,\, = \frac{{6.625 \times {{10}^{ - 34}}}}{{{{10}^{ - 17}}}} \cr
& \,\,\,\,\,\,\,\,\,\, = 6.625 \times {10^{ - 34}} \times {10^{17}} \cr
& \,\,\,\,\,\,\,\,\,\, = 6.625 \times {10^{ - 17}}kg\,m\,{s^{ - 1}} \cr} $$