The de-Broglie wavelength of a proton (mass=1.6×10−27kg) accelerated through a potential difference of 1 kV is :
600˚A
0.9×10−12m
7˚A
0.9×10−19nm
A
600˚A
B
0.9×10−12m
C
7˚A
D
0.9×10−19nm
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Solution
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eV=KE=12mv2
KE=p22m
De-broglie wavelength,
λ=hp
λ=h√2mE=h√2mVe
Putting the values of constants, we get:
λ=6.6×10−34√2×1.6×10−27×1.6×10−19×1000=0.9×10−12m
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