AP EAMCET202422 May 2024Evening ShiftChemistryStructure of AtomActual
The de Broglie wavelength of an electron with kinetic energy of 2.5 eV is (in m ) (1 eV =1.6 10⁻¹⁹ ~J , m_ e =9 10⁻³¹ ~kg )
Options
- Ah 10⁻²⁵ 72
- Bh 10²⁵ 72
- C72 h 10⁻²⁵
- D72 h 10²⁵
Correct answer
B. h 10²⁵ 72
Step-by-step solution
KE = 1 2 ~m v^2v= 2 KE ~m de broglie wavelength ( )= h mv aligned & = h ~m 2 . KE & = h 9 10⁻³¹ 2 2.5 1.6 10⁻¹⁹ & = h 72 10⁻⁵⁰ & = h 72 10²⁵ aligned