For a moving cricket ball,which of the follow | vedclass

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(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ moving with momentum $p$ is given by $\lambda = \frac{h}{p}$.Since kinetic energy $E

Vedclass · Accepted Answer

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ moving with momentum $p$ is given by $\lambda = \frac{h}{p}$.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.While the de-Broglie wavelength exists for all matter,for a macroscopic object like a cricket ball,the wavelength is extremely small (on the order of $10^{-34} \ m$),making it physically undetectable and practically irrelevant,but the mathematical expression remains valid.

For a moving cricket ball,which of the following statements regarding the de-Broglie wavelength is correct?

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