In a photoelectric experiment for \[4000\dot A\] incident radiation, the potential difference to stop the ejection is 2V. If the incident light changed to \[3000\dot A\], then the potential required to stop the ejection of electrons will be:
A. \[2V\]
B. Less than \[2V\]
C. \[Zero\]
D. Greater than \[2V\]

Hint: The minimum amount of energy required for electron emission from the metal surface is the work function 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, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _0}}}\].

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 = incident energy
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }}\] = 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:
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.

Given: In a photoelectric experiment, incident radiation \[{\lambda _1} = 4000\dot A\], the potential difference to stop the ejection is \[{V_{o1}} = 2V\]. The incident light is changed to \[{\lambda _2} = 3000\dot A\]. Speed of light is \[c{\rm{ }} = 3 \times {10^8}m/s\]. We need to determine the new stopping potential.

Equation for photoelectric effect is \[h\nu = {\phi _o} + {E_k}\]---- (1)
\[{\phi _o} = \dfrac{{hc}}{{{\lambda _o}}}\]----(2) and \[{E_k} = e{V_o}\]----- (3)
Also, \[E = h\nu = \dfrac{{hc}}{\lambda }\]--- (4)
Substituting (2), (3) and (4) in (1), we get,
\[\dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{{\lambda _o}}} + e{V_o}\]
\[\Rightarrow e{V_o} = hc(\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}) \\ \]
\[\therefore {V_o} = \dfrac{{hc}}{e}(\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}) \\ \]--- (5)
We know, \[1\dot A = {10^{ - 10}}m\]
So, the stopping potential is inversely proportional to wavelength. As \[{\lambda _1} > {\lambda _2}\], so the new stopping potential will increase. The new stopping potential will be higher than 2V.

Hence option D is the correct answer.

Note: If the energy provided is less than the work function, no electron emission or photoelectric effect will occur. Work function is a metal attribute that is dependent on the metal. The electron is emitted from the surface due to the residual energy after the work function. This additional energy is turned into kinetic energy, allowing the electron to emit from the metal surface. Because wavelength and frequency are inversely related, the stopping potential is proportional to the radiation frequency.