Question

The de Broglies wavelength of a particle having a momentum of 3.3 into 10-24kgms ​

Answer 1

Answer:

Answer: The De-Broglie wavelength of the given particle is 2\times 10^{-10}m2×10

−10

m

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{p}λ=

p

h

where,

\lambdaλ = De-Broglie's wavelength = ?

h = Planck's constant = 6.6\times 10^{-34}Js6.6×10

−34

Js

p = momentum of the particle = 3.3\times 10^{-24}kg.m/s3.3×10

−24

kg.m/s

Putting values in above equation, we get:

\begin{gathered}\lambda=\frac{6.6\times 10^{-34}Js}{3.3\times 10^{-24}kgm/s}\\\\\lambda=2\times 10^{-10}m\end{gathered}

λ=

3.3×10

−24

kgm/s

6.6×10

−34

Js

λ=2×10

−10

m

Hence, the De-Broglie wavelength of the given particle is 2\times 10^{-10}m2×10

−10

m

Explanation:

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