Given:
Frequency of light (v1) = 3.2 x 10^16 Hz
Frequency of light (v2) = 2.0 x 10^16 Hz

To find:
Threshold frequency (v0) of the metal

Formula:
The equation for the photoelectric effect is given by:
hv = φ + 1/2 mv^2

Where,
h = Planck's constant (6.626 x 10^-34 J.s)
v = Frequency of light
φ = Work function
m = Mass of the photoelectron

Let's find the relationship between the kinetic energies (K.E.) of the two photoelectrons.

1. Relationship between K.E. and Frequency:
For the first case (v1), the kinetic energy (K.E.1) is given by:
K.E.1 = hv1 - φ

For the second case (v2), the kinetic energy (K.E.2) is given by:
K.E.2 = hv2 - φ

Given that K.E.2 = 2K.E.1, we can equate the two equations:
hv2 - φ = 2(hv1 - φ)

2. Relationship between Frequency and Threshold Frequency:
We know that the threshold frequency (v0) is the minimum frequency of light required to emit photoelectrons. Below the threshold frequency, no photoelectrons are emitted.

For the first case (v1), if v1 > v0, then photoelectrons are emitted. If v1 < v0,="" then="" no="" photoelectrons="" are="" />

Similarly, for the second case (v2), if v2 > v0, then photoelectrons are emitted. If v2 < v0,="" then="" no="" photoelectrons="" are="" />

3. Solving the Equations:
Using equation (1), we can rewrite equation (2) as:
hv2 - φ = 2hv1 - 2φ

Rearranging the terms, we get:
hv2 - 2hv1 = -φ

Now, let's substitute the equation for frequency (v) in terms of the threshold frequency (v0) for both cases:

For the first case (v1):
v1 = v0 + Δv1
Δv1 represents the difference between v1 and v0.

For the second case (v2):
v2 = v0 + Δv2
Δv2 represents the difference between v2 and v0.

Substituting these values in the rearranged equation, we get:
h(v0 + Δv2) - 2h(v0 + Δv1) = -φ

Expanding the equation, we get:
h(v0 + Δv2 - 2v0 - 2Δv1) = -φ

Simplifying the equation, we get:
h(Δv2 - 2Δv1) = -φ - hv0

Since φ is the work function, it is always positive. Therefore, -φ is negative.

This implies that Δv2 - 2Δv1 is also negative. In other words, the difference between the two frequencies (Δv2) is less than twice the difference between the threshold frequency and the first frequency (2Δv1).

Therefore, the threshold frequency (v0) must be greater than v2 in order