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`

Explanation

`E= -(13.6eV)Z^(2)((1)/((Z+2)^(2))-(1)/(Z^(2)))`
`= -13.6xx((Z^(2)-(Z+2)^(2))/((Z+2)^(2)))=(4(Z+1)xx13.6)/((Z+2)^(2))eV`.....(`1`)
Now energy of electron is `k=(h^(2))/(2lambda^(2)m)`, (we have `lambda=(h)/(sqrt(2mk))`)
or, `k=6eV`
So, `(4(Z+1)xx13.6)/((Z+2)^(2))=6+4.2=10.2eV`
Or, `((Z+1))/((Z+2)^(2))=(3)/(16)rarr(Z-2)(3Z+2)=0`
So, the value of `Z=2` (neglecting the negative/fractional value)