The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
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Solution 1
According to de Broglie’s expression,
`lambda = h/(mv)`
Substituting the values in the expression,
`lambda = (6.626xx10^(-34)Js)/(0.1 kg)(4.37xx10^5 ms^(-1))`
`lambda = 1.516xx 10^(38) m`
Solution 2
λ = h/mv = 6.626×10-34 kgm2s-1/(0.1 kg) (4.37×105 ms-1) = 1.516×10-28 m
Concept: Bohr'S Model for Hydrogen Atom
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