24P2Q07

7a)

KE of a photoelectron is equal to the energy of photon minus the work required to break free from the lattice attraction.

Even though the energy of each photon is the same, the electrons are bound to different degree to the metal lattice.

The liberated photoelectrons hence have a range of KE, with the maximum value belonging to the least tightly bound electron, which is equal to energy of photon minus the work function..

7b)

\displaystyle \begin{aligned}K{{E}_{\max }}&=hf-\Phi  \\3.48e&=3.95e-\Phi  \\\Phi &=4.47\text{ eV}  \end{aligned}
.

\displaystyle \begin{aligned}\Phi &=\frac{hc}{{{\lambda }_{\max }}} \\4.47e&=\frac{(6.63\times {{10}^{-34}})(3.00\times {{10}^{8}})}{{{\lambda }_{\max }}} \\{{\lambda }_{\max }}&=2.78\times {{10}^{-7}}\text{ m}  \end{aligned}

7c)

The number of photoelectrons per unit time hitting metal plate is now higher.

This should lead to more photoelectrons being ejected per unit time.

Time taken to fully discharge is shorter.

7d)

Infrared’s wavelength is too long.

The energy of each photon is lower than the work function.

This means each photon does not contain sufficient energy to liberate even the least tightly bound electron.

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