The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

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#### Solution

No

Work function of the metal, `phi_0 = 4.2 eV`

Charge on an electron, *e *= 1.6 × 10^{−19} C

Planck’s constant, *h* = 6.626 × 10^{−34} Js

Wavelength of the incident radiation, *λ* = 330 nm = 330 × 10^{−9} m

Speed of light, *c* = 3 × 10^{8} m/s

The energy of the incident photon is given as:

`E = "hc"/lambda`

`= (6.626 xx 10^(-34) xx 3 xx 10^8)/((330 xx 10^(-9))) = 6.0 xx 10^(-19) J`

`= (6.0 xx 10^(-19))/(1.6 xx 10^(-19)) = 3.76 eV`

It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.

Concept: Einstein’S Photoelectric Equation: Energy Quantum of Radiation

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