Question

If work function of a photoelectric material is $3.3eV$, then threshold frequency will be:
$A)8\times {{10}^{14}}Hz$
$B)8\times {{10}^{56}}Hz$
$C)8\times {{10}^{10}}Hz$
$D)8\times {{10}^{11}}Hz$

Answer 1

Hint: Work function is defined as the minimum energy required to remove an electron from the surface of a solid to vacuum. It is calculated in $J$ or $eV$. Work function of a photoelectric material is proportional to the threshold frequency of electromagnetic radiation.
Formula used:
$\phi =h{{f}_{0}}$

Complete answer:
Work function of a photoelectric material is defined as the minimum energy required to remove an electron from the surface of the photoelectric material to a point outside the photoelectric material. Threshold frequency is defined as the minimum frequency of electromagnetic radiation required to remove an electron from the surface of a photoelectric material. Therefore, it is clear that both threshold frequency of electromagnetic radiation as well as work function of a photoelectric material are proportional to each other. Mathematically, work function of a photoelectric material is given by
$\phi =h{{f}_{0}}$
where
$\phi $ is the work function of a photoelectric material
${{f}_{0}}$ is the threshold frequency of electromagnetic radiation falling on the photoelectric material
$h$ is the Planck’s constant
Let this be equation 1.
From the question, we are provided that the work function of a photoelectric material is equal to $3.3eV$. We are required to find the threshold frequency of electromagnetic radiation used in the photoelectric phenomenon of this material.
Substituting the value of work function in equation 1, we have
$\phi =h{{f}_{0}}\Rightarrow 3.3eV=(6.6\times {{10}^{-34}})\times {{f}_{0}}\Rightarrow {{f}_{0}}=\dfrac{3.3\times 1.6\times {{10}^{-19}}J}{6.6\times {{10}^{-34}}Js}=0.8\times {{10}^{15}}{{s}^{-1}}=8\times {{10}^{14}}Hz$
Therefore, the minimum frequency or the threshold frequency of electromagnetic radiation required to remove an electron from the surface of the given photoelectric material is given by
${{f}_{0}}=8\times {{10}^{14}}Hz$

Hence, the correct answer is option $A$.
Additional information:
Photoelectric effect is defined as the phenomenon of removal of electrons from the surface of a material when electromagnetic radiation is allowed to fall on the surface of the material. Photons in the electromagnetic radiation interact with the electrons on the surface of the photoelectric material, to emit photoelectrons from the surface of the material.

Note:
Students need to be thorough with conversion formulas. The conversion formula used in the above solution is given as follows.
$1J=1.6\times {{10}^{-19}}eV$
It is important to keep in mind that converting units to the SI system of units is the easiest as well as the safest way for calculations.
Students need to remember the value of Planck’s constant too. Planck’s constant is given by
$h=6.6\times {{10}^{-34}}Js$
Also, in the last step of the calculation, ${{s}^{-1}}$ is equated to $Hz$. Students need not get confused because $1Hz$ is defined as one cycle per second$(1{{s}^{-1}})$.