If the kinetic energy of an electron is 100 eV, what is the wavelength of the de-Broglie wave associated with it? [h = 6.6 x 10\(^{-34}\) J s, e = 1.6 x 10\(^{-19}\)C, m\(_{e}\), = 9.1 x 10\(^{-31}\) kg]
A
1.22 x 10\(^{-10}\)m
B
4.10 x 10\(^{-10}\)m
C
3.90 x 10\(^{-10}\)m
D
5.50 x 10\(^{-14}\)m
Verified Correct Answer: A.
Physical Principles & Numerical Solution
de broglie wavelength formula in terms of kinetic energy:
\lambda = \(\frac{h}{\sqrt{2mE}}\)
\lambda = \(\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times10^{-31} \times100\times 1.6 \times 10^{-19}}}\)
\lambda = 1.22 x 10\(^{-10}\)m
NOTE; E = 100eV = 100 x 1.6 x 10\(^{-19}\)J
Source: WAEC / JAMB / NECO Classroom Archive (Record #63368)