Question

An electron is moving with a velocity of 2.5 x 10^6 m/s. lf the uncertainty in its velocity is 0.1%. Calculate the uncertainty in its position.

Answer 1

Given,

• Velocity of electron (v) = 2.5 × 10⁶ ms⁻¹

• Uncertainity (Δv) = 0.1 % = 0.001

• Quantum Number (n) = 4

We know,

• Planck's constant (h) = 6.623 × 10⁻³⁴ Js

• Mass of electron (m$_e$) = 9.1 × 10⁻³¹ kg

Δv = Uncertainity × Velocity

⇒ Δv = 2.5 × 10⁶ × 0.001

⇒ Δv = 2500 ms⁻¹

According to Uncertainity principle,

$\boxed{\displaystyle\sf{\Delta x\:.\:\Delta v=\frac{h}{2\pi m_e}}}$

The Uncertainity in Position,

$\boxed{\displaystyle\sf{\Delta x=\frac{h}{2\pi}\times\frac{1}{m_e\Delta v}}}$

$\implies\displaystyle\sf{\Delta x=\frac{6.626\times 10^{-34}\times 1}{4\times 3.14\times 9.1\times 10^{-31}\times 2500}}$

$\implies\displaystyle\sf{\Delta x=\frac{6.626\times 10^{-34}}{285740\times 10^{-31}}}$

$\implies\displaystyle\sf{\Delta x=2.3\times 10^8\:m}$

$\boxed{\displaystyle\sf{Uncertainty \:in \:position=2.3\times 10^8\:m}}$

NOTE :-

• As the value of n varies in above equation, the value the Uncertainity position is also varies.

• You may use the scientific calculator for calculations.

Answer 2

Answer:

ᴀᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ʜᴇɪsᴇɴʙᴇʀɢ's ᴜɴᴄᴇʀᴛᴀɪɴɪᴛʏ ᴘʀɪɴᴄɪᴘʟᴇ,

ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴇʟᴇᴄᴛʀᴏɴ, v = 2.5 ×10⁶ ms⁻¹

ᴘʟᴀɴᴄᴋ's ᴄᴏɴsᴛᴀɴᴛ, ʜ=6.623×10⁻³⁴ ᴊs

ᴍᴀss ᴏғ ᴇʟᴇᴄᴛʀᴏɴ, ᴍᴇ=9.1×10⁻³¹ kg

Δv = 2.5 * 10⁶ * 0.1 / 100 = 2500 ms⁻¹

=> ᴜɴᴄᴇᴛᴀɪɴɪᴛʏ ᴘᴏsɪᴛɪᴏɴ ᴡʜᴇɴ ɴ = 4 ɪs,

Δx = ʜ/ɴπ * 1/ ᴍᴇ*Δᴠ

= 6.626 * 10⁻³⁴/4π * 1/ 9.1×10⁻³¹ * 2500

= 2.3 * 10⁻⁸ ᴍᴇᴛʀᴇ

=> ᴀs ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ "ɴ" ᴠᴀʀɪᴇs ɪɴ ᴀʙᴏᴠᴇ ᴇϙᴜᴀᴛɪᴏɴ, ᴛʜᴇ ᴠᴀʟᴜᴇ ᴛʜᴇ ᴜɴᴄᴇʀᴛᴀɪɴɪᴛʏ ᴘᴏsɪᴛɪᴏɴ ɪs ᴀʟsᴏ ᴠᴀʀɪᴇs.

Explanation:

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