
In a photoelectric experiment the stopping potential for the photoelectron is \[2V\] for the incidence of wavelength \[400nm\]. If the incidence the light is changed to \[300nm\], the cut-off potential is
A) \[2V\]
B) Greater than\[\dfrac{8}{3}V\]
C) \[\dfrac{8}{3}V\]
D) Zero
Answer
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Hint: When the light is illuminated in the metals, the photons are emitted. This process of emission of electrons on the illumination of light is called the photoelectric effect. The emitted electrons are called photoelectrons. The photons impart its energy to remove the electron from its orbit and also for its motion after emitting. Hence the energy of the photon is used for removing the electrons and also for its kinetic energy.
Complete step by step answer:
(i) When the negative end of the power supply is connected to the photoelectric material, it sets up potential differences across the ends. If the low voltage sets on it, the least number of electrons can be emitted. The electrons emitted will be restricted.
(ii) According to the potential the work function of the metal behaves. Hence,
\[ \Rightarrow P = \dfrac{{hc}}{\lambda }\]
Where,
\[P\] is the potential work function
\[h\] is the Planck’s constant and the value is $6.626 \times {10^{ - 34}}J.s$
$c$ is the velocity of the light and the value is $3.8 \times {10^8}\dfrac{m}{s}$
$\lambda $ is the wavelength of the light.
The product of the Planck’s constant and the velocity divided by the wavelength of the light is known as the work potential.
(iii) When the incident light of wavelength is 400nm and the potential is 2V,
\[hc = 2 \times 400nm\]
(iv)When the incidence wavelength is 300nm, the cut-off potential changes in to,
\[ \Rightarrow P = \dfrac{{2 \times 400}}{{300}}\]
\[ \Rightarrow P = \dfrac{{800}}{{300}}\]
\[\therefore P = \dfrac{8}{3}V\]
Hence the cut-off potential or stopping potential for the 300 nm wavelength of light is \[\dfrac{8}{3}V\]. Therefore option (C) is correct.
Note:
The cut-off potential is also called stopping potential. The stopping potential controls the emission of photoelectrons removed from the metal. This is set up in the power supply. This potential determines the work function of the photoelectric material. The photoelectric effect was discovered by Albert Einstein in 1905.
Complete step by step answer:
(i) When the negative end of the power supply is connected to the photoelectric material, it sets up potential differences across the ends. If the low voltage sets on it, the least number of electrons can be emitted. The electrons emitted will be restricted.
(ii) According to the potential the work function of the metal behaves. Hence,
\[ \Rightarrow P = \dfrac{{hc}}{\lambda }\]
Where,
\[P\] is the potential work function
\[h\] is the Planck’s constant and the value is $6.626 \times {10^{ - 34}}J.s$
$c$ is the velocity of the light and the value is $3.8 \times {10^8}\dfrac{m}{s}$
$\lambda $ is the wavelength of the light.
The product of the Planck’s constant and the velocity divided by the wavelength of the light is known as the work potential.
(iii) When the incident light of wavelength is 400nm and the potential is 2V,
\[hc = 2 \times 400nm\]
(iv)When the incidence wavelength is 300nm, the cut-off potential changes in to,
\[ \Rightarrow P = \dfrac{{2 \times 400}}{{300}}\]
\[ \Rightarrow P = \dfrac{{800}}{{300}}\]
\[\therefore P = \dfrac{8}{3}V\]
Hence the cut-off potential or stopping potential for the 300 nm wavelength of light is \[\dfrac{8}{3}V\]. Therefore option (C) is correct.
Note:
The cut-off potential is also called stopping potential. The stopping potential controls the emission of photoelectrons removed from the metal. This is set up in the power supply. This potential determines the work function of the photoelectric material. The photoelectric effect was discovered by Albert Einstein in 1905.
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