NTA Abhyas JEE Main2020ChemistryStructure of AtomPractice
The de-Broglie wavelength of an α -particles at a voltage V is (Given that α -particle has 2 units positive charge and 4 units mass)
Options
- Aλ= 12 .3 V A o
- Bλ= 0 .286 V A o
- Cλ= 0 .101 V A o
- Dλ= 0 .856 V A o
Correct answer
C. λ= 0 .101 V A o
Step-by-step solution
The de-Broglie wavelength associated with the charged particle as λ= h mv 1 2 mv 2 = K .E 1 2 mv 2 × 2m = 2m K .E = m 2 v 2 2m· K .E = mv λ = h 2 m K.E = h 2 m α · 2 eV = 6.626 × 1 0 - 3 4 2 × 1.66 × 1 0 - 2 7 × 4 × 1.6 × 1 0 - 1 9 × 2 × V = 6.626 × 1 0 - 3 4 V × 1 0 - 2 3 × 6.519 = 1.018 × 1 0 - 1 1 V = 0 .1018 V   A o