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What  is the de Broglie wavelength associated with an electron moving with a speed of  \(v=5.4\times 10^6\;\mathrm{m/s}\).

Answer

\(13\mathrm{\ldotp }5\times {10}^{−11}\;\mathrm{m}\)
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Solution

For the electron:
Mass, \(m=9.11\times 10^{-31}\;\mathrm{kg}\)
 speed  \(v=5.4\times 10^6\;\mathrm{m/s}\)
Then momentum,  
 \(p=mv\) 
 \(=9.11\times 10^{-31}\times 5.4\times 10^6\) 
 \(=4.92\times 10^{-24}\;\mathrm{kgm/s}\) 
According to the de Broglie equation,
 \(\lambda =\frac h p\) 
Planck’s constant,  \(h=6.62\times 10^{-34}\mathrm{Js}\)   
Substituting both the values for calculating de Broglie’s wavelength,
 \(\lambda =\frac{\left(6.62\times 10^{-34}\right)}{4.92\times 10^{-24}}\) 
 \(=13.5\times 10^{-11}\;\mathrm{m}\)
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