Uncertainty in position of an electron (mass of an electron is =$9.1\times10^{-28}g)$ moving with a velocity of $3 \times 10 ^4$ cm/s accurate upto 0.001% will be (use $\frac{h}{4\pi}$ in uncertainty expression where $h = 6.626 \times 10^{-27}$ erg s)

Question

An element 'M' has 25% of the electrons filled in the third shell as in the second shell. The element 'M' is :

Answer 1

To solve this problem, we will use Heisenberg's Uncertainty Principle, which states that it's impossible to simultaneously determine the exact position and momentum of a particle, such as an electron. The principle is mathematically represented as:

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

where \Delta x is the uncertainty in position, \Delta p is the uncertainty in momentum, and h is the Planck's constant.

Given:

The mass of the electron m = 9.1 \times 10^{-28} \text{ g}
The velocity of the electron v = 3 \times 10^4 \text{ cm/s}
The accuracy of measurement in velocity is 0.001%, implying that the uncertainty in velocity \Delta v = 0.001\% \times v
Planck's constant h = 6.626 \times 10^{-27} \text{ erg s}

Let's calculate the uncertainty in velocity:

\Delta v = \frac{0.001}{100} \times 3 \times 10^4 = 3 \times 10^{-4} \text{ cm/s}

The uncertainty in momentum \Delta p is given by:

\Delta p = m \cdot \Delta v = 9.1 \times 10^{-28} \cdot 3 \times 10^{-4} = 2.73 \times 10^{-31} \text{ g cm/s}

Now, using the uncertainty principle:

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

Substitute the values we have:

\Delta x \geq \frac{6.626 \times 10^{-27}}{4\pi \cdot 2.73 \times 10^{-31}}

Calculate \Delta x:

\Delta x \geq \frac{6.626 \times 10^{-27}}{4 \times 3.14159 \times 2.73 \times 10^{-31}} \approx 1.93 \text{ cm}

Hence, the uncertainty in the position of the electron is approximately 1.93 cm.

Therefore, the correct answer is: 1.93 cm.

Answer 2