1.

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is (radius of the first orbit of hydrogen atom =0.53Å

Answer»

`1.67Å`
`3.33Å`
`1.06Å`
`0.53Å`

Solution :ACCORDING to Bohr.s quantisationof angular momentum
`mvr=(NH)/(2pi)`
or `(h)/(MV)=(2pir)/(n) ""...(i)`
de-Broglie wavelength
`lambda=(h)/(mv) ""...(ii)`
From EQS (i) and(ii), we get
Wavelength `lambda=(2pir)/(n)`
`=(2pxx PI xx0.53Å)/(1)`
`=3.33Å`


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