
If in photoelectric experiment, the wavelength of incident radiation is reduced from \[6000\mathop A\limits^ \circ \] to \[4000\mathop A\limits^ \circ \] then
A. Stopping potential will decrease
B. Stopping potential will increase
C. Kinetic energy of emitted electron will decrease
D. The value of the work function will decrease
Answer
161.1k+ views
Hint: When a photon strikes metal, it carries enough energy to overcome the attractive attraction that holds the valence electron to the shell of the metal's atom. If the photon's energy exceeds the minimum energy required to evict the electron, the remainder of the energy is transmitted as kinetic energy of the ejected electrons.
Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \],
where K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
Complete step by step solution:
The metal's threshold wavelength is the wavelength that corresponds to the minimal energy required to overcome the attractive attraction that holds the valence electron to the shell of the metal's atom. Because photon energy is inversely related to wavelength, the wavelength should be the highest permitted wavelength for a lowest energy value.
The stopping potential is the potential which is able to stop the ejected electron. Using the work energy theorem, when a body is acted upon by external force then the change in kinetic energy of the body is equal to the work done on it.
The work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi \\ \]
\[\Rightarrow {V_0} = \dfrac{{K{E_{\max }}}}{e} = \dfrac{{hc}}{{e\lambda }} - \phi \]
As the stopping potential is inversely proportional to the wavelength of the light, so when we decrease the wavelength from \[6000\mathop A\limits^ \circ \] to \[4000\mathop A\limits^ \circ \] the stopping potential will increase. The kinetic energy of the ejected electron will increase. The work function is the characteristic property of the metal, it is independent of the incident light’s wavelength. So, it remains the same as the initial.
Therefore, the correct option is B.
Note: As we know, frequency is inversely related to wavelength. As a result, the threshold frequency is the lowest value of the photon's frequency that contains enough energy to evict the electron from the metal.
Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \],
where K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
Complete step by step solution:
The metal's threshold wavelength is the wavelength that corresponds to the minimal energy required to overcome the attractive attraction that holds the valence electron to the shell of the metal's atom. Because photon energy is inversely related to wavelength, the wavelength should be the highest permitted wavelength for a lowest energy value.
The stopping potential is the potential which is able to stop the ejected electron. Using the work energy theorem, when a body is acted upon by external force then the change in kinetic energy of the body is equal to the work done on it.
The work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }} = \dfrac{{hc}}{\lambda } - \phi \\ \]
\[\Rightarrow {V_0} = \dfrac{{K{E_{\max }}}}{e} = \dfrac{{hc}}{{e\lambda }} - \phi \]
As the stopping potential is inversely proportional to the wavelength of the light, so when we decrease the wavelength from \[6000\mathop A\limits^ \circ \] to \[4000\mathop A\limits^ \circ \] the stopping potential will increase. The kinetic energy of the ejected electron will increase. The work function is the characteristic property of the metal, it is independent of the incident light’s wavelength. So, it remains the same as the initial.
Therefore, the correct option is B.
Note: As we know, frequency is inversely related to wavelength. As a result, the threshold frequency is the lowest value of the photon's frequency that contains enough energy to evict the electron from the metal.
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