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Question: The maximum velocity of the photoelectron emitted by the metal surface is \(v\) . Charge and mass of...

The maximum velocity of the photoelectron emitted by the metal surface is vv . Charge and mass of the photoelectron is denoted by ee and mm respectively. The stopping potential in volt is?
A. v22(me)\dfrac{{{v^2}}}{{2\left( {\dfrac{m}{e}} \right)}}
B. v22(em)\dfrac{{{v^2}}}{{2\left( {\dfrac{e}{m}} \right)}}
C. v2(em)\dfrac{{{v^2}}}{{\left( {\dfrac{e}{m}} \right)}}
D. v2(me)\dfrac{{{v^2}}}{{\left( {\dfrac{m}{e}} \right)}}

Explanation

Solution

Hint The relation between maximum kinetic energy of photoelectron and stopping potential is given by the equation 12mvmax2=eV0\dfrac{1}{2}mv_{\max }^2 = e{V_0} where mm is the mass of photoelectron, vmax{v_{\max }} is its maximum velocity and V0{V_0} is the stopping potential.

Complete step-by-step solution :
Einstein and Millikan are the two physicist who described the photoelectric effect using a formula that is the relation between the maximum kinetic energy (Kmax)\left( {{K_{\max }}} \right) of the photoelectrons and the frequency of the absorbed photons (ν)\left( \nu \right) and the threshold frequency (ν0)\left( {{\nu _0}} \right) of the photoemissive surface.
Kmax=h(νν0){K_{\max }} = h\left( {\nu - {\nu _0}} \right) where hh is plank’s constant.
The maximum kinetic energy (Kmax)\left( {{K_{\max }}} \right) of the photoelectrons (with charge e) can be determined from the stopping potential as
V0=Wq=Kmaxe\Rightarrow {V_0} = \dfrac{W}{q} = \dfrac{{{K_{\max }}}}{e}. So,
Kmax=eV0\Rightarrow {K_{\max }} = e{V_0}
On further simplifying we have
12mvmax2=eV0\Rightarrow \dfrac{1}{2}mv_{\max }^2 = e{V_0} where mm is the mass of photoelectron, vmax{v_{\max }} is its maximum velocity and V0{V_0} is the stopping potential.
As given in the question, the maximum velocity of the photon vmax=v{v_{\max }} = v. So,
12mv2=eV0\Rightarrow \dfrac{1}{2}m{v^2} = e{V_0}
On simplifying we have
V0=v22(em)\Rightarrow {V_0} = \dfrac{{{v^2}}}{{2\left( {\dfrac{e}{m}} \right)}}
Hence, option B is correct.

Note: Stopping potential states that the potential required to stop the emission of electrons from a metal surface when a beam of light having energy greater than the work potential of metal is incident on it.
When this negative voltage increases, it prevents all the highest-energy electrons from reaching the collector. When there is no current passing through the tube it means the negative voltage has reached the maximum value to slow down and stop the most energetic photoelectrons of kinetic energy Kmax{K_{\max }} . This negative value of the retarding voltage is called the stopping potential or cut off potential V0{V_0}. Since the work done by the retarding potential in stopping the electron of charge e is eV0e{V_0} , the following formula must hold Kmax=eV0{K_{\max }} = e{V_0}