Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohr's orbit `n = 2`, returns to the orbit `n = 1`, in the hydrogen atom .The ionisation potential of the ground state hydrogen atom is `2.17 xx 10^(-11)` ergs per atom

Answer

A, B, D

Explanation

`E_(2) - E_(1) = - 2.17 xx 10^(-11) [(1)/(4) - (1)/(1)] erg [n_(1) = 1, n_(2) = 2]`
`= 2.17 xx 10^(-11) xx (3)/(4) erg`
`E_(2) - E_(1) = hv = (c )/(lambda) h`
`lambda = (hc)/(Delta E ) = (6.62 xx 10^(-27) erg xx 3.0 xx 10^(10) cm )/(2.17 xx 10^(-11) xx 3//4)`
` = (2.648)/(2.17 xx 10^(-7) cm )= 1.22 xx 10^(-5) cm = 1220 Å`