A beam of monochromatic light of wavelength `lambda` ejects photoelectrons from a cesium `(phi = 1.9eV)` these photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of `lambda` for which
(a) hydrogen atoms may be ionised
(b) hydrogen may get excited from the ground state to the first excited state and
(c ) the excited hydrogen atoms may emit visible light

Answer

(a) `lambda=(hc)/((13.6+1.9)eV)=80 nm`
(b) `lambda=(hc)/((10.2+1.9)eV)=102 nm`
(c ) `(hc)/((12.08+1.9)eV)=89nm`

Explanation

(a) To ionize the `H` atom in ground minimum `K.E`. Of photoelectron needed `=13.6 eV`. And `because phi =1.9 eV`
`rArr` Minimum energy (or maximum wavelength) incident `=13.6+1.9=15.5 eV`
or `lambda_(max)=(6.6xx10^(34)xx3xx10^(8))/(15.5xx1.6xx10^(-19))80nm`
(b) `DeltaE= 13.6 ((1)/(1)(1)/(2^(2)))`
`=(3xx13.6)/(4)=10.2 eV=` minimum `K.E.` of photoelectron neede and `because phi=1.9 eV`
`rArr` Minimum energy (or maximum wavelength) incident `=10.2+1.9=12.1 eV`
or `lambda_(max)=(6.6xx10^(-34)xx3xx10^(8))/(12.1xx1.6xx10^(-19))=102nm`
(c ) Any transition to `n=2` from higher state in a `H`-atom results in the emission of visible light. So `H`-atom the minimum `K.E`. of photoelectron needed `=E_(3)-E_(1)= -15.4+13.6=12.08 eV` and ` because phi=1.9eV`
minimum energy incident `(hc)/(lambda_(max))=12.08+phi=12.08+1.98 eV`
`:. lambda_(max)=(6.6xx10^(-34)xx3xx10^(8))/(13.98xx1.6xx10^(-19))=89 nm`