A gas containining hydrogen-like ions, with atomic number `Z`, emits photons in transition `n + 2 to n, where n = Z` These photons fall on a metallic plate and eject electrons having minimum de Broglie wavelength `lambda of 5 Å`. Find the value of `Z` , if the work function of metal is `4.2 eV`

Answer

2

Explanation

`E = - (13.6 e V) Z^(2) ((1)/((Z + 2)^(2)) - (1)/(Z^(2)))`
`= - 13.6 xx ((Z^(2) - (Z + 2)^(2))/((Z + 2)^(2))) = (4( Z + 1) xx 13.6)/((Z + 2)^(2)) eV` (i)
[Now energy of electron is `K = (h^(2))/(2 lambda^(2) m)`
Solving `K = 6 eV` ltrbgt ` (4( Z + 1) xx 13.6)/((Z + 2)^(2)) = 6 + 4.2 = 10.2 eV`
`( Z + 1)/(((Z + 2)^(2)) = (3)/(16) implies (Z - 2) (3 Z + 2) = 0`
So , the value of `Z = 2` (neglecting the negative//fractional value)