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Given that

The electron is freely in a closed cubic box (CCB) of length $(a)=10cm$

Also given that

The electron is freely moving inside the box.

Therefore,

The maximum uncertainly in position will be equal to the length of the diagonal.

$Δx=3a $

So,

$Δx=3 ×10$

$Δx(mΔv)=4πh $

$Δv=4π(m)Δxh $

$⇒103 ×10_{3}×9.1×10_{−31}×4×3.146.623×10_{−34} $

$⇒5×10_{−4}ms_{−1}$

Hence,

The maximum uncertainty in its velocity $5×10_{−4}ms_{−1}$Solve any question of Structure of Atom with:-

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