what is the de broglie wavelength of an electron with kinetic energy of 1200 ev
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Answer:
3.546×10^-11m
Explanation:
As P = √( 2m× KE)
= √ ( 2× 9.1×10 ^-31× 1200× 1.6×10^-19)
Where ,
Mass of the electron = 9.1× 10^ -31 kg
And 1 eV= 1.6 × 10 ^ -19
P= √[3.4944×10^-46]
= 1.869×10^-23 Kgm/s
Then
de-Broglie wavelength
lamda = h/p
h is Planck constant : 6.63 ×10^-34
Lamda = 6.63×10^-34 ÷ 1.869×10^-23
= 3.546× 10^-11 m
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