Question

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 1

Wavelengths corresponding to minimum wavelength (λ_min ) or maximum energy will emit photoeiectrons having kinetic energy which is maximum. (λmin ) belonging to Balmer series and Lying in the given range (450 nm to 750 nm) corresponds to transition from (n = 4 to n = 2). Here,
$\left( E \right)_{4} = - \frac{1 3 \cdot 6}{\left(4\right)^{2}} = - 0 \cdot 8 5 \, \text{eV}$
$\left(\text{and} E \right)_{2} = - \frac{1 3 \cdot 6}{\left(2\right)^{2}} = - 3 \cdot 4 \, \text{eV}$
$\therefore \Delta E = E ⁡_{4} - E ⁡_{2} = 2 \cdot 5 5 \, \text{eV}$
$ K _{\text{max}} = \text{Energy of photon} - \text{work function} = 2 \cdot 5 5 - 2 \cdot 0 = 0 \cdot 5 5 \text{eV}$