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Question
If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms-1, calculate the de Broglie wavelength associated with it.
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Solution
According to de Broglie’s equation,
`lambda = "h"/("mv")`
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ:
`lambda = (6.626 xx 10^(-34) " Js")/((9.10939 xx 10^(-31)" kg")(2.19xx10^6 " ms"^(-1))`
`= 3.32 xx 10^(-10)" m" = 3.32 xx 10^(-10) " m" xx 100/100`
`= 332xx10^(-12) " m"`
λ = 332 pm
∴ Wavelength associated with the electron = 332 pm
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