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In a hydrogen atom the de-Broglie wavele...

In a hydrogen atom the de-Broglie wavelength of an electron is `1.67nm`. The value of principal quantum number of the electron is

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Verified by Experts

The correct Answer is:
5
`lamda=(2pir_(n))/n`
`lamda=(2xx3.14xx0.529xx10^(-10)xxn^(2))/n`
`1.67xx10^(-9)=2xx3.14xx0.529xx10^(-10)xxn`
`:.n=5`
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