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Question: Radiation of \(\lambda \)=155 nm was irradiated on Li (work function=5 eV) plate. The stopping poten...

Radiation of λ\lambda =155 nm was irradiated on Li (work function=5 eV) plate. The stopping potential (in eV) is:
A. 3 eV
B. 8 eV
C. 9 eV
D. 5 eV

Explanation

Solution

We know that stopping potential is the difference of voltage needed to stop the movement of electrons between plates in a photoelectric experiment. Here, we have to use the formula of stopping potential, that is, V0=E−work function{V_0} = E - {\rm{work}}\,{\rm{function}}. Here, V0{V_0}
is stopping potential and E is kinetic energy.

Complete step by step answer:
Let’s first calculate the Kinetic energy of the radiation. The formula to calculate energy of radiation is,
Energy of radiation=hcλ\dfrac{{hc}}{\lambda } …… (1)
Here, h represents Planck’s constant, c is speed of light and λ\lambda is wavelength.
The value of h is 6.626×10−34 J s6.626 \times {10^{ - 34}}\,{\rm{J}}\,{\rm{s}} and value of c is 3×108 m s−13 \times {10^8}\,{\rm{m}}\,{{\rm{s}}^{ - 1}}.
Given the value of wavelength is 155 nm. We have to convert wavelength into metre. So, the value of λ\lambda =155×10−9 m155 \times {10^{ - 9}}\,{\rm{m}}
As we need the value of stopping potential in eV, we have to multiply 1.6×10−191.6 \times {10^{ - 19}}
to the denominator of equation (1). So, the formula becomes,
Energy of radiation=hcλ×1.6×10−19\dfrac{{hc}}{{\lambda \times 1.6 \times {{10}^{ - 19}}}}
Now, we have to put the values of c, h and λ\lambda to the above formula.
Energy of radiation=6.626×10−34×3×108 155×10−9×1.6×10−19=8.015 eV\dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}\,}}{{155 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} = 8.015\,{\rm{eV}}
Now, we have to calculate the stopping potential using the following formula, that is,
V0=E−work function{V_0} = E - {\rm{work}}\,{\rm{function}}
The energy of radiation is 8.015 eV and the work function is 5 eV. So, the stopping potential is,
V0=8.015 eV−5 eV=3.015 eV{V_0} = 8.015\,{\rm{eV}} - {\rm{5}}\,{\rm{eV}} = {\rm{3}}{\rm{.015}}\,{\rm{eV}}

So, the correct answer is Option A.

Note: Always remember that photoelectric effect is the phenomenon in which electrons are ejected from the surface of certain metals on striking the surface with light radiations. Max Planck was the first scientist who gave the relation between frequency of the radiation and radiation associated with them, that is, E=hvE = hv