Question

A silver sphere of radius $1cm$ and work function $4.7eV$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200nm$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A\times {{10}^{z}}$ $( where, 1 < A < 10 )$. The value of ‘z’ is
A. 6
B. 7
C. 8
D. 9

Answer 1

The minimum energy required to extract one electron from a material is known as the work function of that material. When light higher than the work function of a metal is concentrated over its surface the electrons of the material get excited and emit from the surface. As the electrons emitted on the surface of the metal will acquire a potential due to the loss of electrons.
As per the given data,
Work function ($\phi$) is $4.7eV$
The wavelength of the light $200nm$
No of the photoelectrons emitted $A\times {{10}^{z}}\quad (1< A < 10)$

Formula used:
$E=h\upsilon$
$V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{r}$

Complete step by step answer:
Considering a spherical body of $4.7eV$ which is illuminated with light of wavelength $200nm$. The total energy of the system will be the sum of the minimum energy required to extract an electron (work function) and the voltage difference required to stop the flow of the electrons (stopping potential).
Mathematically,
$E=\phi +e{{V}_{0}}$
As we know that according to Einstein’s photoelectric equation,
$E=h\upsilon$
So here we can say that the total energy of the spherical system will be
$h\upsilon =\phi +e{{V}_{0}}$
By rearranging
$e{{V}_{0}}=\dfrac{h\upsilon }{\lambda }-\phi$
So, by putting the values the stopping potential of the system will be,
$\begin{aligned} & e{{V}_{0}}=\dfrac{1240}{200}-4.7 \\\ & \Rightarrow e{{V}_{0}}=6.2-4.7 \\\ & \therefore e{{V}_{0}}=1.5V \\\ \end{aligned}$
So the stopping potential of the sphere is $1.5V$.
When the electrons are emitted the potential acquired (stopping potential) by the sphere is given by,
$V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{r}$
Where,
$r$ is the radius of the sphere
(This value of potential is derived by the gauss law from electrostatics)
This can be also written as,
$\begin{aligned} & V=k\left( \dfrac{ne}{r} \right) \\\ & \Rightarrow 1.5=9\times {{10}^{9}}\left( \dfrac{n\times 1.6\times {{10}^{-19}}}{{{10}^{-2}}} \right) \\\ \end{aligned}$
By rearranging the no of electrons emitted is given as,
$\begin{aligned} & n=1.5\left( \dfrac{9\times {{10}^{-9}}\times {{10}^{2}}}{1.6\times {{10}^{-19}}} \right) \\\ & \therefore n=1.2\times {{10}^{7}} \\\ \end{aligned}$
Thus by comparing, the maximum no electrons emitted by the sphere is,
$n=1.2\times {{10}^{7}}=A\times {{10}^{z}}$
The value of z will be equal to 7.

Thus the correct option which satisfies the question is Option B.

Note:
When there is an emission of a maximum no of electrons a stopping potential is gained by the sphere. After acquiring this there will be no flow of the electrons and no gain of potential. The total energy of such a system is the sum of the work function and the stopping potential of the material.