Determine the maximum kinetic energy of the photoelectrons emitted when light of wavelength 350 nm strikes a cesium surface. The work function of cesium is 1.9 eV .

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 1.6 eV

The energy of the incident photon is given by

E = \dfrac{hc}{\lambda}

Using

hc \approx 1240 \text{ eV nm}

and

\lambda = 350 \text{ nm}

:

E = \dfrac{1240}{350} \approx 3.54 \text{ eV}

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is:

K_{\text\max} = E - \Phi

Substituting the values of

E

and the work function

\Phi = 1.9 \text{ eV}

:

K_{\text\max} = 3.54 \text{ eV} - 1.9 \text{ eV} = 1.64 \text{ eV}

The closest value among the given options is

1.6 \text{ eV}

.

Determine the maximum kinetic energy of the photoelectrons emitted...

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