Question

In the photoelectric effect, it is found that incident photons with energy 5.00ev will produce electrons with a maximum kinetic energy 3.00ev. What is the threshold frequency of this material?

Answer 1

Answer:

The threshold frequency of this material is $4.83\times 10^{14}\ Hz$

Explanation:

It is given that,

Energy of incident photon, E = hυ = 5 eV

Maximum kinetic energy of electrons,  $E_k=3\ eV$

We have to find the threshold frequency of this material. Einstein photoelectric equation gives the relationship between energy and threshold frequency i.e.

$h\nu=h\nu_0+E_k$............(1)

where

$\nu_0$ is the threshold frequency of this material.

h is Planck's constant, $h=4.1357\times 10^{-15}\ eV\ s$

Putting all values in equation (1) as :

$5\ eV=h\nu_0+3\ eV$

$\nu_0=\dfrac{2\ eV}{4.1357\times 10^{-15}\ eV\ s}$

$\nu_0=4.83\times 10^{14}\ Hz$

Hence, the threshold frequency of this material is $4.83\times 10^{14}\ Hz$                                                                             

Answer 2

From Literature:

The amount of energy in the photons is given by this equation:

E = hf

where E = energy
            h = Planck's constant = 6.63 * 10^-34 Joule seconds
            f = frequency of the light, Hz

Given:

E= 3.00 eV and Planck's constant 

To solve for the frequency, E = 3.00 eV

1 electronvolt = 1.60218 x 10^-19 Joules

3 * 1.60218 x 10^-19 Joules = 6.63 * 10^-34 Joule seconds * f 
f = 7.25 x 10^14 /second or hertz

Therefore, the threshold frequency of the material is 7.25 x 10^14 Hertz.