Question

A particle of mass 6.6×10^-31kg is moving with a velocity of 1×10⁷ms^-1. The de Broglie wavelength (in A) associated with the particle, is (h=6.6×10^-34 Js)

• A. 1
• B. 10
• C. 5
• D. 2
• E. 4

Answer 1

The Correct Option is A

The de Broglie wavelength \( \lambda \) is given by the formula: \[ \lambda = \frac{h}{mv} \] where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity. Substituting the given values: \[ \lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-31} \times 1 \times 10^7} \] \[ \lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-24}} = 10^{-10} \, \text{m} = 1 \, \text{Å} \] Thus, the de Broglie wavelength associated with the particle is \( 1 \, \text{Å} \).

The correct option is (A) : \(1\)

Answer 2

Given:
Mass of the particle, m = 6.6 × 10^-31 kg
Velocity of the particle, v = 1 × 10⁷ m/s
Planck's constant, h = 6.6 × 10^-34 Js

de Broglie wavelength formula: 
\(\lambda = \frac{h}{mv}\)

Substitute the values:
\(\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-31} \times 10^7}\)
\(\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-24}} = 10^{-10} \text{ m}\)

Convert meters to angstroms:
\(1 \, \text{Å} = 10^{-10} \, \text{m}\)
So, \(\lambda = 1 \, \text{Å}\)

✅ Correct answer: 1