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?
Answer & explanation
Correct answer: option 1
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,
$K_{max}=E_{photon}-\phi$
$=(2.55-1.9)eV$
$=0.65eV$