Answer 1

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 × 10⁸ 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.