In hydrogen atom light corresponding to the transition n = 4 to n = 2 falls on cesium metal. The work function of the metal is 1.9 eV. What is the maximum kinetic energy of the photoelectrons emitted?

0.65 eV

The correct answer is Option (1) → 0.65 eV

The energy levels in the hydrogen atom is -

$E_n=-13.6×\frac{1}{n^2}$

for the transition $n=4→n=2$,

$E_{photon}=E_4-E_2=\left(-13.6×\frac{1}{4^2}\right)-\left(-16.3×\frac{1}{2^2}\right)$

$=2.55eV$

According to photoelectric equation,

$K_{max}=E_{photon}-\phi$

$=(2.55-1.9)eV$

$=0.65eV$