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.

Answer 2

The kinetic energy of the electrons is 2.1  × 10^-19 J while the velocity of the electron is 6.78 × 10^5 m/s.

The work function of the metal is obtained as follows;

Wo = hfo

Wo = work function

h = Plank's constant

fo =  threshold frequency

Wo = 6.63 × 10^-34 Js ×  5.46 × 10^14 s-1 = 3.6 ×  × 10^-19 J

The energy of the incident photon is obtained from;

E = hc/λ

E = energy of incident photon

h = Plank's constant

c = speed of light

λ = wavelength of photon

E = 6.63 × 10^-34 Js × 3 × 10^8/350 × 10^-9

E = 5.7  × 10^-19 J

From the Einstein equation;

KE = E - Wo

KE = 5.7  × 10^-19 J -  3.6 ×  10^-19 J = 2.1  × 10^-19 J

Since;

KE = 1/2mv^2

v = √2KE/m

v = √2 × 2.1  × 10^-19 J/9.11 × 10^-31kg

v = 6.78 × 10^5 m/s

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