For what energy a proton will the associated de-broglie wavelength be 16.5 nm
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Answer: kinetic energy
Explanation:
Here, λ=16.5nm=16.5×10−9m
K.E. of proton,
K=12(mv2)=12m(mv2)=12mh2λ2
=(6.63×10−34)22×(1.675×10−27)×(16.5×10−9)2
=4.82×10−25J
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