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

Explanation

(a) Ionisation energy for H-atom=13.6 eV
`E=phi+13.6=1.9+13.6=15.5 eV`
`lambda=(1242)/(E(eV))nm =(1242)/(15.5)=80 nm`
Excitation energy for n=1 to n=2 is
`13.6((1)/(1^(2))-(1)/(2^(2)))=10.2 eV`
`E=phi +10.2=1.9+10.2=12.1 eV`
`lambda=(1242)/(E(eV))nm=(1242)/(12.1)=102.6 nm`
Excitation energy for n =1 to n=3 is
`13.6((1)/(1^(2))-(1)/(3^(2)))=12.1 eV`
n=3 to n=2, visible light
`E=phi+12.1=1.9+12.1=14 eV`
`lambda=(1242)/(E(eV))nm=(1242)/(14)=88.7 nm`