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Uncertainty in position of an electron (mass = 9.1 × 10–28 g) moving with a velocity of 3 × 104 cm/s accurate upto 0.001% will be (use h/4π) in uncertainty expression where h  = 6.626 ×10–27 erg-second) [1995]
  • a)
    1.93 cm
  • b)
    3.84 cm
  • c)
    5.76 cm
  • d)
    7.68 cm
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Uncertainty in position of an electron (mass = 9.1 × 10–28...
1. **Given data**
- Mass of electron (m) = 9.1 × 10^(-28) g
- Velocity of electron (v) = 3 × 10^4 cm/s
- Accuracy = 0.001%
- Planck's constant (h) = 6.626 × 10^(-27) erg-second
2. **Calculating uncertainty in position**
The uncertainty in position (∆x) can be calculated using the formula:
∆x = h / (4π * m * ∆v)
Where:
- h = Planck's constant
- m = mass of the electron
- ∆v = uncertainty in velocity
3. **Calculating uncertainty in velocity**
Uncertainty in velocity can be calculated as a percentage of the actual velocity:
∆v = (0.001/100) * v
∆v = 0.001 * 3 × 10^4
∆v = 30 cm/s
4. **Substitute values in the formula**
∆x = (6.626 × 10^(-27)) / (4π * 9.1 × 10^(-28) * 30)
∆x = 1.93 cm
5. Therefore, the uncertainty in the position of the electron moving with a velocity accurate up to 0.001% is 1.93 cm. Hence, option **A** is correct.
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Community Answer
Uncertainty in position of an electron (mass = 9.1 × 10–28...
Let uncertainity in position be X
then,
x • m∆v = h/4 Π
x= h/4Π×m∆v
now put the values
x= 6.626×10^-27/4×3.14×9.1×10^-28×3.4×10^4
so by solving it we will get
x= 1.931×10^-5
now, according to the question uncertainity is correct upto to the 0.001%
So, 0.001% = 1.931×10^-5
Therefore, x= 1.93 cm
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Uncertainty in position of an electron (mass = 9.1 × 10–28 g) moving with a velocity of 3 × 104 cm/s accurate upto 0.001% will be (use h/4π) in uncertainty expression where h = 6.626 ×10–27 erg-second) [1995]a)1.93 cmb)3.84 cmc)5.76 cmd)7.68 cmCorrect answer is option 'A'. Can you explain this answer?
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