Question

Light of wavelength \[4000\dot A\] falls on a photosensitive metal and a negative 2 V potential stops the emitted electrons. The work function of the material (in eV) is approximately:
A. \[2\]
B. \[2.2\]
C. \[3.1\]
D. \[1.1\]

Answer 1

Hint:From the equation for photoelectric effect \[h\nu = {\phi _o} + {E_k}\] , the work function is defined as the minimum amount of energy required for electron emission from the metal surface. It is the characteristic property of that metal and the frequency of light corresponding to this minimum energy is called threshold frequency and the corresponding wavelength is called threshold wavelength. It is calculated as a work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _0}}}\]. \[{E_k}\] is the energy of the emitted electron. It is also known as the maximum kinetic energy such that \[{E_k} = e{V_o}\] where \[{V_o}\] is the stopping potential.

Formula(e) used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
Maximum kinetic energy,
\[{E_k} = e{V_o}\]
Where,
E = energy of incident radiation
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }} = \text{speed of light} = 3 \times {10^8}\,m/s\]
\[\nu \]= frequency of the light,
\[{\nu _o}\]= threshold frequency,
\[\lambda \]= wavelength of the light,
\[{\lambda _o}\]= threshold wavelength
\[e = 1.6 \times {10^{ - 19}}C\]= electronic charge
\[{V_o}\]= stopping potential

Complete step by step solution:
Given: In this photoelectric experiment, wavelength of light is \[\lambda = 4000\dot A\] and a negative stopping potential is \[{V_o} = 2{\rm{ }}V\]or 2eV. We need to determine the work function of this metal.
\[c{\rm{ }} = \text{speed of light}= 3 \times {10^8}m/s\]
\[\Rightarrow e = 1.6 \times {10^{ - 19}}C\]

Equation for photoelectric effect is,
\[h\nu = {\phi _o} + {E_k} \\ \]
\[\Rightarrow \dfrac{{hc}}{\lambda } = {\phi _o} + {E_k} \\ \]---- (1)
\[\Rightarrow {\phi _o} = \dfrac{{hc}}{{{\lambda _o}}} \\ \]----(2)
\[\Rightarrow {E_k} = e{V_o}\]---- (3)
From equation (1),
\[{\phi _o} = \dfrac{{hc}}{\lambda } - {E_k}\]---(4)
Substituting equation (3) in (2), we get,
\[{\phi _o} = \dfrac{{hc}}{\lambda } - e{V_o} \\ \]--- (5)
\[\Rightarrow 1\dot A = {10^{ - 10}}m\]

Energy of incident radiation,
\[E = \dfrac{{hc}}{\lambda } = \dfrac{{(6.64 \times {{10}^{ - 34}})(3 \times {{10}^8})}}{{4000 \times {{10}^{ - 10}}}}\]
Solving this will give the value of energy in joules.
The value of energy of incident radiation in eV is:
\[E = \dfrac{{(6.64 \times {{10}^{ - 34}})(3 \times {{10}^8})}}{{4000 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}} \\ \]
\[\Rightarrow E = 3.1\,eV\]
Using equation (5) and substituting all values in it, we get,
\[{\phi _o} = 3.1 - 2\]
\[\therefore {\phi _o} = 1.1\,eV\]

Hence option D is the correct answer.

Additional information: Photoelectric effect was discovered by Einstein in 1905 for which he also won the Nobel prize in physics. According to its theory, when a light of sufficient energy is incident on a metal surface, the photons of the incident light impart energy to the electrons on the metallic surface. If this energy is higher than the threshold energy of the metal, the electrons become sufficiently energetic to escape the metal surface and are emitted. These electrons are called photoelectrons whose energy is less than the energy of the incident light as some of the energy is utilised in overcoming the barrier energy or the work function.

Note: Work function is a property of the metal that depends on the metal. The remaining energy after the work function is converted to maximum kinetic energy to enable electron emission from the metal surface. Also, the minimum negative voltage needed to be applied at the anode is the stopping potential and is found equal to this maximum kinetic energy when converted in electron volts.