1.

If energy of an electron is 11.375 eV, then its de-Broglie wavelength associated with the moving elec-tron is(1) 32.3 Å(2) 40.5 Å(3) 36 4 Å(4) 3

Answer»

LAMDA= h/√(2mE)= (6.626×10^-34)/√(2×9.1×10^-31×11.375×1.6×10^-19)=3.6406×10^-10 m=3.6406 A°



Discussion

No Comment Found

Related InterviewSolutions