Question

The threshold frequency for metallic potassium is 5.46×1014s-1 calculate the maximum kinetic energy and velocity that the emitted electron has the wave length of light shining on the potassium surface is 350nm(Me=9.11×1031kg)

Answer 1

Given :

The threshold frequency for metallic potassium is 5.46×1014 s⁻¹.

To Find :

The maximum kinetic energy and velocity that the emitted electron has the wave length of light shining on the potassium surface is 350 nm.

Solution :

We know, Maximum kinetic energy is given by :

$K.E = \dfrac{hc}{\lambda}-h\nu_o\\\\K.E = h \times( \dfrac{3\times 10^8}{350\times 10^{-9}} -5.46\times 10^{14} )\\\\K.E = 6.626 \times 10^{-34}\times ( \dfrac{3\times 10^8}{350\times 10^{-9}} -5.46\times 10^{14} )\\\\K.E = 2.06 \times 10^{-19}\ J$

Hence, this is the required solution.