Question

Calculate the value of work function in $eV$ .
(A) $1.77$
(B) $3.27$
(C) $5.69$
(D) $2.32$

Answer 1

The work function is defined as the minimum amount of energy needed to eject an electron from the surface of the metal particle to infinity. Work function is denoted by the Greek symbol Phi ( $\Phi \;$ ). It is usually expressed in $eV$ (electron volts) or $J$ units (Joules).

Formula Used: We will use the following formula to calculate work function
$W = \dfrac{{hc}}{{{\lambda _0}}}$
Where $W$ is the work function,
$h$ is the Planck’s constant,
$c$ is the speed of light,
${\lambda _0}$ is the threshold wavelength.

Complete step by step solution:
According to the question the given threshold wavelength ${\lambda _0} = 3800{\mathop A\limits^\circ} = 3800 \times {10^{ - 10}}m$
Also, we know that
$h = 6.62 \times {10^{ - 34}}$
And
$c = 3 \times {10^8}m/s$
Now we will put all the known values in the work function formula
Then we will get
$W = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3800 \times {{10}^{ - 10}}}}$
$\Rightarrow W = 5.23 \times {10^{ - 19}}J$
So, we got our answer in terms of joules. But we need to convert joules to electron volts.
We know that
$1eV =$ $1.6 \times {10^{ - 19}}\;J$
$\therefore 5.23 \times {10^{ - 19}}J = 3.27eV$
Our required answer is $3.27eV$ .
Hence the correct answer is option B.

Note:
Work functions rely on a surface's structure and chemical composition. Different crystallographic surfaces of the same metal or compound, for instance, can have significantly different work functions. A surface's chemical modifications can have even larger consequences. Some electrons, very close to the surface, gain sufficient energy to overcome the working function of the material and are emitted from the surface with kinetic energy when light energy of sufficient intensity strikes the surface of the material. These electrons which are emitted are called photo-electrons.
We will solve this question for the answer in joules. At the end, make sure to convert the final answer to electron volts as asked in the question.