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:

Answer 1

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:
$V={\displaystyle \frac{kq}{R}}$
$U=eV$
$U={\displaystyle \frac{hc}{\lambda }}-\varphi$

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, ${\varphi }_o$
Kinetic energy of photons, $K.E$
Equation of photoelectric effect:
$E={\varphi }_o+K.E$
$h\upsilon =h{\upsilon }_o+{\displaystyle \frac{1}{2}}m{v}^2$
Let $x$ be the maximum number of photoelectrons emitted.
The potential $V$ of the sphere can be calculated as,
$V={\displaystyle \frac{kq}{R}}$
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,
$V={\displaystyle \frac{9\times {10}^9\times x\times 1.6\times {10}^{-19}}{{10}^{-2}}}$
$V=x\times 14.4\times {10}^{-8}$
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,
$U={\displaystyle \frac{hc}{\lambda }}-\varphi$
Where,
$U$ is the kinetic energy of emitted photons
${\displaystyle \frac{hc}{\lambda }}$ is the energy of the incident light
$\varphi$ is the work function of the metal surface
Putting all the values,
$$x\times 23.04\times {10}^{-27}={\displaystyle \frac{2\times {10}^{-25}}{200\times {10}^{-9}}}-4.7\times 1.6\times {10}^{-19}$$
$x={\displaystyle \frac{{10}^{-18}-7.52\times {10}^{-19}}{2}}$
$x=1.076\times {10}^7$
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.