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

Options

A. 1.67 Å
B. 3.33 Å
C. 1.06 Å
D. 0.53 Å

Answer

A

Explanation

De-Broglie wavelength of a particle is given by
`lamda=(h)/(mv)` . . . (i)
Where, h is planck's constant.
If kinetic energy of particle of mass m is v, then
`K=(1)/(2)mv^(2) implies v=sqrt((2K)/(m))` . . . (ii)
Combining Eqs. (i) and (ii) we get
`lamda=(h)/(msqrt((2K)/(m)))=(h)/(sqrt(2mK))` . . . (iii)
Given, `m=9.1xx10^(-31)kg`
`K=10keV=10xx10^(3)xx1.6xx10^(-19)J`
`h=6.6xx10^(-34)J-s`
Substituting the above values in Eq. (iii), we get
`lamda=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx10xx10^(3)xx1.6xx10^(-19)))`
`=1.22xx10^(-11)~~0.12` Å