Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work function is 2.0 eV. Find ( in eV ) the maximum kinetic energy of the emitted photoelectrons. `("Take hc = 1242 eV nm.")`

Answer

0.55

Explanation

Wavelengths corresponding to minimum wavelength `(lamda_(min))` or maximum energy will emit photoelectrons having kinetic energy which is maximum . (`lamda` min) belonging to Balmer series and Lying in the given range (450 nm to 750 nm) correspondig to transition from (n=4 to n=2). Here
`E_(4)=-13.6/((4)^(2))=-0.85eV`
and `E_(2)=-13.6/((2)^(2))=-3.4eV`
`:.DeltaE=E_(4)-E_(2)=2.55eV`
`K_("max")=` Energy of photon-work function =2
`.55-2.0=0.55eV`