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:

Hint: When an electromagnetic radiation of particular wavelength and frequency falls on a metal surface, electrons are emitted having a particular value of kinetic energy. We will apply the equation of photoelectric effect to find the number of electrons emitted when light of given wavelength falls on the surface of a silver sphere.

Formulae used:
$V=\dfrac{kq}{R}$
$U=eV$
$U=\dfrac{hc}{\lambda }-\phi $

Complete step by step answer:
The photoelectric effect is described as the emission of electrons when electromagnetic radiation such as light hits a metal surface material. Electrons emitted during this process are called Photoelectrons. We can say that in photoelectric effect, electrically charged particles, such as electrons, are released from or within a material when it absorbs electromagnetic radiation. This process is described as the ejection of electrons from a metal surface plate when light falls on it.

Terms related to photoelectric effect:
Energy of the incident light,$E$
Work function of the metal, ${{\phi }_{o}}$
Kinetic energy of photons, $K.E$
Equation of photoelectric effect:
$E={{\phi }_{o}}+K.E$
$h\upsilon =h{{\upsilon }_{o}}+\dfrac{1}{2}m{{v}^{2}}$
Let $x$ be the maximum number of photoelectrons emitted.
The potential $V$ of the sphere can be calculated as,
$V=\dfrac{kq}{R}$
Where,
$k$ is the Coulomb’s law constant
$q$ is the charge acquired by the particle
$R$ is the radius of the body
Putting values of $k,q$ and $R$, we get,
$V=\dfrac{9\times {{10}^{9}}\times x\times 1.6\times {{10}^{-19}}}{{{10}^{-2}}}$
$V=x\times 14.4\times {{10}^{-8}}$
Energy of photons is given by,
$U=eV$
$U=x\times 14.4\times {{10}^{-8}}\times 1.6\times {{10}^{-19}}$
$U=x\times 23.04\times {{10}^{-27}}$
From conservation of energy, we have,
$U=\dfrac{hc}{\lambda }-\phi $
Where,
$U$ is the kinetic energy of emitted photons
$\dfrac{hc}{\lambda }$ is the energy of the incident light
$\phi $ is the work function of the metal surface
Putting all the values,
\[x\times 23.04\times {{10}^{-27}}=\dfrac{2\times {{10}^{-25}}}{200\times {{10}^{-9}}}-4.7\times 1.6\times {{10}^{-19}}\]
$x=\dfrac{{{10}^{-18}}-7.52\times {{10}^{-19}}}{2}$
$x=1.076\times {{10}^{7}}$
Number of photoelectrons emitted $=1.076\times {{10}^{7}}$
Comparing with the given expression,
$A\times {{10}^{z}}$$=1.076\times {{10}^{7}}$
Therefore,
$z=7$

Note: Light waves of any frequency cannot cause photoelectrons to be emitted. Light below a certain cut off frequency will cause the photoelectric effect. For light above the cut off voltage; the more intense the light, the higher kinetic energy the emitted photons would have. For a constant value of intensity of light, the number of photons emitted is inversely proportional to the frequency of incident light.