Solveeit Logo
Login

Question

Chemistry Question on Structure of atom

The momentum of a particle having a de-Broglie wavelength of 101710^{17} m is (Given, h=6.625×10−34h = 6.625 \times 10^{-34} m)

A

3.3125×10−7kgms−13.3125 \times 10^{-7} kg m s^{-1}

B

26.5×10−7kgms−126.5 \times 10^{-7} kg m s^{-1}

C

6.625×10−17kgms−16.625 \times 10^{-17} kg m s^{-1}

D

13.25×10−17kgms−113.25 \times 10^{-17} kg m s^{-1}

Answer

6.625×10−17kgms−16.625 \times 10^{-17} kg m s^{-1}

Explanation

Solution

According to de-Broglie relation,
where, λ\lambda = wavelength
             \, \, \, \, \, \, \, \, \, \, \, \, \, h = Planck's constant
             \, \, \, \, \, \, \, \, \, \, \, \, \, p = momentum
Here,          h=6.625×10−34Js\, \, \, \, \, \, \, \, \, h = 6.625 \times 10^{-34} Js
              \, \, \, \, \, \, \, \, \, \, \, \, \, \, \lambda=10−17m=10^{-17}m
∴      p=hλ6.625××10−3410−17\therefore \, \, \, \, \, \, p=\frac{h}{\lambda}\frac{6.625\times \times10^{-34}}{10^{-17}}
                        =6.625×10−34×1017\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 6.625 \times10^{-34} \times 10^{17}
                        =6.625×10−17kgms−1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 6.625 \times 10^{-17} kg m s^{-1}