Question

AIIMS level question help karo ab !!!​

Answer 1

Answer:

$\huge\underline\orange{\mathcal Answer}$

$\huge\blue{\mathcal (a)2}$

According To Heisenberg Uncertainty Principle :-

$\large{\delta x×m{\delta v={\frac{h}{4π}}}}$

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For Particle A :-

$\large{\delta x ={\delta Xa}}$

$\large{ m=m}$

$\large{\delta v =0.05}$

therefore,

$\large{\delta Xa × m×0.02={\frac{h}{4π}}}-->(1)$

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For Particle B :-

$\large{\delta x ={\delta Xb}}$

$\large{m =5m}$

$\large{\delta v =0.02}$

therefore ,

$\large{\mathcal{\delta Xb×5m×0.02={\frac{h}{2π}}}}-->(2)$

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By equation (1) & (2)

$\huge{\boxed{\mathcal{\frac{{\delta Xa}}{{\delta Xb}}=2}}}$

$\huge\orange{\mathcal Hope\:it\:helps!!!}$