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Question:

Calculate the kinetic energy of the electron having wavelength 1nm.

4.2 eV

2.1 eV

3.1 eV

1.5 eV

Solution:

De-Broglie wavelength, λ=h/mv
λ²=h²/m²v²
λ²=h²/2m(K.E)
K.E.=h²/2mλ²=(6.6×10⁻³⁴)²/2×9.1×10⁻³¹×(10⁻⁹)²=2.42×10⁻¹⁹ J=2.42×10⁻¹⁹/1.6×10⁻¹⁹ eV ≈1.5 eV