Answer 1

Frequency of the incident photon, `v = 488 nm = 488 xx 10^(-9) m`

Maximum speed of the electrons, v = 6.0 × 10⁵ m/s

Planck’s constant, h = 6.626 × 10⁻³⁴ Js

Mass of an electron, m = 9.1 × 10⁻³¹ kg

For threshold frequency ν₀, the relation for kinetic energy is written as:

`1/2 mv^2 = h(v - v_0)`

`v_0 = v - (mv^2)/(2h)`

= `7.21 xx 10^14 - ((9.1 xx 10^(-31) )xx (6xx10^(5))^2)/(2xx(6.626 xx 10^(-34)))`

Therefore, the threshold frequency for the photoemission of electrons is `4.738 xx  10^14 Hz`