Given below are two statements : Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle. Statement (II) : If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is \geqslant \sqrt\frach\pi \times \frac12 m . In the light of the above statements, choose the correct answer from the options given below :

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Structure of Atom PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q16 with Solution

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Correct answer: B

Check Statement (I)

Statement (I) says:

It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle.

This is exactly Heisenberg's uncertainty principle:

\Delta x \, \Delta p \geq \frac{h}{4\pi}

So, position and momentum cannot both be measured exactly at the same time.

Therefore, Statement (I) is true.

Check Statement (II)

Given:

uncertainty in position = uncertainty in momentum
for an electron, find uncertainty in velocity

Let

\Delta x = \Delta p

Using Heisenberg's uncertainty principle:

\Delta x\,\Delta p \geq \frac{h}{4\pi}

Since $\Delta x = \Delta p$ ,

(\Delta p)^2 \geq \frac{h}{4\pi}

Taking square root,

\Delta p \geq \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}

Now for an electron,

\Delta p = m\,\Delta v

So,

m\,\Delta v \geq \frac{1}{2}\sqrt{\frac{h}{\pi}}

Hence,

\Delta v \geq \sqrt{\frac{h}{\pi}}\times \frac{1}{2m}

This matches Statement (II).

Therefore, Statement (II) is also true.

Conclusion

Statement (I): True
Statement (II): True

So the correct option is:

\boxed{\text{B}}

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View written solutionFree

Correct answer: B

Check Statement (I)

Statement (I) says:

It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle.

This is exactly Heisenberg's uncertainty principle:

\Delta x \, \Delta p \geq \frac{h}{4\pi}

So, position and momentum cannot both be measured exactly at the same time.

Therefore, Statement (I) is true.

Check Statement (II)

Given:

uncertainty in position = uncertainty in momentum
for an electron, find uncertainty in velocity

Let

\Delta x = \Delta p

Using Heisenberg's uncertainty principle:

\Delta x\,\Delta p \geq \frac{h}{4\pi}

Since $\Delta x = \Delta p$ ,

(\Delta p)^2 \geq \frac{h}{4\pi}

Taking square root,

\Delta p \geq \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}

Now for an electron,

\Delta p = m\,\Delta v

So,

m\,\Delta v \geq \frac{1}{2}\sqrt{\frac{h}{\pi}}

Hence,

\Delta v \geq \sqrt{\frac{h}{\pi}}\times \frac{1}{2m}

This matches Statement (II).

Therefore, Statement (II) is also true.

Conclusion

Statement (I): True
Statement (II): True

So the correct option is:

\boxed{\text{B}}

Structure of Atom question

Structure of Atom question

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