Question

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5V$ when the radiation of $250nm$ is used. The work function of the metal is:
A.$5.5$$eV$
B.$4$ $eV$
C.$5$ $eV$
D.$4.5$$eV$

Answer 1

Hint: The stooping energy of a photoelectron is given and hence photons don’t have mass but they energy. The energy of every photon is inversely proportional to its wavelength. Determining the energy of a photon will help in knowing the work function of the photon.

Formula Used:
To solve the above numerical we need different formulas, which are given below:
$E = \dfrac{{hc}}{\lambda }$
Here $E$ is the energy of a photon.
$h$ is Planck constant
$c$ Speed of light
\[\lambda \] is wavelength of a photon
Kinetic energy = Stopping potential
$E$= $W$+ $K.E.$
Here $E$is Energy of photon
$W$ is work function
$K.E.$ is Kinetic energy

Complete step by step solution:
For solving the above numerical we need to follow steps given below:
Given radiation that is wavelength
\[\lambda \]= $250$ $nm$
\[\lambda \]= $2500$\[{A^o}\] (as $1$$nm$ = $10$\[{A^o}\]
Using energy formula for photon, we will get Energy value
$E = \dfrac{{hc}}{\lambda }$
= $\dfrac{{12400}}{{2500}}$ (as value of $hc$= $12400$ \[{A^o}\] )
= $4.96$$eV$
Since stopping potential is given. So,
Kinetic Energy = stopping potential = $0.5$$eV$
Now to determine the work function
$E$= $W$+ $K.E.$
$4.96$= $W$+ $0.5$ ($E$=$4.96$, $K.E.$= $0.5$)
$W$= $4.46$\[ \approx \]$4.5$$eV$

So the correct answer is option D.

Additional Information:
Photoelectric Effect is emission of electrons when radiation like light hits on material or surface. The electrons which got emitted are known as Photoelectrons. This theory is given by Einstein for which he won the Nobel Prize in Physics. This theory is basically important for understanding the quantum nature of light and electrons.
Photon Energy is defined as energy which is carried by a single photon. Energy is directly proportional to photon’s electromagnetic frequency and inversely proportional to wavelength. Higher the frequency, higher the energy of the photon. However high wavelengths lower the energy of photons.
Unit of Photon energy is electron volt or joule. Different ways or formula of photon are there depends on what things are provided in numerical, we can select that equation:
$E = \dfrac{{hc}}{\lambda }$
Here $E$ is the energy of a photon.
$h$ is Planck constant
$c$ Speed of light
\[\lambda \] is wavelength of a photon
Since $\dfrac{c}{\lambda } = f$, here it denotes frequency
So now,
$E$= $hf$
This equation is known as Planck-Einstein relation.
Also we can write this equation as:
$E$= $h\nu \;$
Here $h$ is Planck’s constant
And \[\nu \;\] is the photon's frequency.

Note: Photon’s energy, work function etc. all depend upon either wavelength or frequency. To determine the energy of a photon or the work function of a photon, we should know about the concept of wavelength or frequency.