An electron in H-atom in its ground state absorbs ` 1.5` times as much energy as the minimum required for its escape ( i. e., 13 . 6 eV) from the atom . Calculate the wavelength of emitted electron.

Explanation

The energy required to remove an electron
from H atom = Ionisation energy of H atom
`= 2.18 xx 10^(-18) J "atom"^(-1)`
The amount of energy absorbed = `1.50 xx 2.18 xx 10^(-18) J`
`= 3.27 xx 10^(-18) J`
Amount of energy converted to kinetic energy
`=3.27 xx 10^(-18) - 2.18 xx 10^(-18) J`
`= 1.09 xx 10^(-18) J`
K.E. = `1/2 mv^2`
`v = sqrt((2 K.E)/(m)) = sqrt((2 xx (1.09 xx 10^(-18)J))/((9.1 xx 10^(-31) kg)))`
`=1.55 xx 10^(6) ms^(-1)`
According to de-Broglie equation,
`lambda = h/(mv)`
`= (6.63 xx 10^(-34) Js)/((9.1 xx 10^(-31) kg) xx (1.55 xx 10^(6) ms^(-1)))`
`= 4.70 xx 10^(-10) m`.