
A photon of frequency n causes photoelectric emission from a surface with threshold frequency \[{v_0}\]. The de Broglie wavelength \[\lambda \] of the photoelectron emitted is given as:
A. \[\Delta n = \dfrac{h}{{2m\lambda }}\]
B. \[\Delta n = \dfrac{h}{\lambda }\]
C. \[\left[ {\dfrac{1}{{{v_0}}} - \dfrac{1}{v}} \right] = \dfrac{{m{c^2}}}{h}\]
D. \[\lambda = \sqrt {\dfrac{h}{{2m\Delta n}}} \]
Answer
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Hint: The threshold energy defines the minimum kinetic required by the reactant molecules for the effective collision to give the product. Only the molecules that possess energy greater than or equal to threshold energy, take part in chemical reactions.
Complete Step by Step Answer:
Let's discuss the photoelectric effect in detail. The exposure of a metal surface to light energies such as X rays, gamma rays, etc, of suitable frequency, causes the emission of electrons. And this phenomenon is termed the photoelectric effect.
The formula of threshold energy is,
\[{E_T}\] =Work function or threshold energy + kinetic energy of the electron
We know, \[E = hv\] , here, h stands for Planck's constant and v stands for frequency.
So, the above equation becomes,
\[hv = h{v_o} + \dfrac{1}{2}m{u^2}\] …… (1)
Here, m stands for the mass of the electron, and u is the speed of the molecule.
On rearranging the equation (1), we get,
\[\dfrac{1}{2}m{u^2} = h\left( {v - {v_0}} \right)\]
We know that, \[v - {v_0} = \Delta n\]
Therefore, the above equation becomes,
\[\dfrac{1}{2}m{u^2} = h\Delta n\]…… (2)
Also, we know, \[\lambda = \dfrac{h}{{mu}},u = \dfrac{h}{{m\lambda }}\]
On putting the above values in equation (2), we get,
\[\dfrac{1}{2}m \times \dfrac{{{h^2}}}{{{m^2}{\lambda ^2}}} = h\Delta n\]
\[\lambda = \sqrt {\dfrac{h}{{2m\Delta n}}} \]
Therefore, option D is true.
Note: It is to be noted that the work function defines the minimum energy requirement of an electron to get emitted from a metal surface. The photoelectrons constituted a current called photoelectric current. Non-metals also show the photoelectric effect. But, gases and liquid show this effect up to a limited extent.
Complete Step by Step Answer:
Let's discuss the photoelectric effect in detail. The exposure of a metal surface to light energies such as X rays, gamma rays, etc, of suitable frequency, causes the emission of electrons. And this phenomenon is termed the photoelectric effect.
The formula of threshold energy is,
\[{E_T}\] =Work function or threshold energy + kinetic energy of the electron
We know, \[E = hv\] , here, h stands for Planck's constant and v stands for frequency.
So, the above equation becomes,
\[hv = h{v_o} + \dfrac{1}{2}m{u^2}\] …… (1)
Here, m stands for the mass of the electron, and u is the speed of the molecule.
On rearranging the equation (1), we get,
\[\dfrac{1}{2}m{u^2} = h\left( {v - {v_0}} \right)\]
We know that, \[v - {v_0} = \Delta n\]
Therefore, the above equation becomes,
\[\dfrac{1}{2}m{u^2} = h\Delta n\]…… (2)
Also, we know, \[\lambda = \dfrac{h}{{mu}},u = \dfrac{h}{{m\lambda }}\]
On putting the above values in equation (2), we get,
\[\dfrac{1}{2}m \times \dfrac{{{h^2}}}{{{m^2}{\lambda ^2}}} = h\Delta n\]
\[\lambda = \sqrt {\dfrac{h}{{2m\Delta n}}} \]
Therefore, option D is true.
Note: It is to be noted that the work function defines the minimum energy requirement of an electron to get emitted from a metal surface. The photoelectrons constituted a current called photoelectric current. Non-metals also show the photoelectric effect. But, gases and liquid show this effect up to a limited extent.
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