Question 45

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the emwave ? (Given hc = 1242eVnm where h is the Planck's constant and c is the speed of light in vaccum.)

$$ K_{max} = e V_0 $$

$$ K_{max} = e(2 \text{ V}) = 2 \text{ eV} $$

Einstein's photoelectric equation

$$ E = \Phi + K_{max} $$

$$ E = 2.14 \text{ eV} + 2 \text{ eV} $$

$$ E = 4.14 \text{ eV} $$

The energy of the incident photon:

$$ E = \frac{hc}{\lambda} $$

$$ \lambda = \frac{hc}{E} $$

$$ \lambda = \frac{1242 \text{ eV nm}}{4.14 \text{ eV}} $$

$$ \lambda = 300 \text{ nm} $$

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