Passage IIThe position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nmQ.Uncertainty in momentum isa)1.3 x10-28 kg ms-1b)1.3 x 10-26kg ms-1c)2.6 x 10-26kg ms-1d)2.6 x 10-28kg ms-1Correct answer is option 'C'. Can you explain this answer?

Answer:

To determine the uncertainty in momentum, we need to use the Heisenberg uncertainty principle, which states that the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) must be greater than or equal to a certain value, given by Planck's constant divided by 4π.

Heisenberg Uncertainty Principle:
Δx * Δp ≥ h/4π

Given:
Energy of the electron = 0.5 keV
Position uncertainty (Δx) = 0.2 nm

Step 1: Converting Energy to Momentum:
The energy of the electron can be converted to momentum using the equation:
E = p^2 / (2m)
where E is the energy, p is the momentum, and m is the mass of the electron.

Since the mass of the electron is known (9.1 x 10^-31 kg), we can rearrange the equation to solve for momentum:
p = √(2mE)

Substituting the values, we get:
p = √(2 * 9.1 x 10^-31 kg * 0.5 x 10^3 eV)

Step 2: Converting Momentum to SI Units:
To convert the momentum to SI units, we need to convert the electronvolt (eV) to joules (J). Since 1 eV = 1.6 x 10^-19 J, we can multiply the momentum by this conversion factor.

p = √(2 * 9.1 x 10^-31 kg * 0.5 x 10^3 eV) * (1.6 x 10^-19 J / 1 eV)

Step 3: Calculating the Uncertainty in Momentum:
Using the Heisenberg uncertainty principle equation, we have:
Δx * Δp ≥ h/4π

Rearranging the equation to solve for Δp, we get:
Δp ≥ h/(4πΔx)

Substituting the values, we get:
Δp ≥ (6.63 x 10^-34 J s) / (4π * 0.2 x 10^-9 m)

Simplifying the equation, we get:
Δp ≥ 1.32 x 10^-26 kg m/s

Therefore, the uncertainty in momentum is 1.32 x 10^-26 kg m/s, which corresponds to option 'C'.

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Passage IIThe position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nmQ.Uncertainty in momentum isa)1.3 x10-28 kg ms-1b)1.3 x 10-26kg ms-1c)2.6 x 10-26kg ms-1d)2.6 x 10-28kg ms-1Correct answer is option 'C'. Can you explain this answer?

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Passage IIThe position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nmQ.Uncertainty in momentum isa)1.3 x10-28 kg ms-1b)1.3 x 10-26kg ms-1c)2.6 x 10-26kg ms-1d)2.6 x 10-28kg ms-1Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Passage IIThe position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nmQ.Uncertainty in momentum isa)1.3 x10-28 kg ms-1b)1.3 x 10-26kg ms-1c)2.6 x 10-26kg ms-1d)2.6 x 10-28kg ms-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIThe position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nmQ.Uncertainty in momentum isa)1.3 x10-28 kg ms-1b)1.3 x 10-26kg ms-1c)2.6 x 10-26kg ms-1d)2.6 x 10-28kg ms-1Correct answer is option 'C'. Can you explain this answer?.

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