Question

Energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength \[2000\mathop A\limits^o \]falls on the surface, the velocity of the fastest electron ejected from the surface will be
A. \[8.4 \times {10^5}{\rm{ m/sec}}\]
B. \[7.4 \times {10^5}{\rm{ m/sec}}\]
C. \[6.4 \times {10^5}{\rm{ m/sec}}\]
D. \[8.4 \times {10^6}{\rm{ m/sec}}\]

Answer 1

Hint: The least amount of energy necessary to cause photoemission of electrons from a metal surface is known as the work function. The combination of the energy used to expel an electron and the maximal kinetic energy of electrons makes up the total energy of a photon. The electron's velocity may be determined by applying this idea to get kinetic energy.

Formula used The energy of the photon is given by the equation is given as:
\[E = h\upsilon \].
Where \[h\] is the Plank constant and \[\upsilon \] is the frequency of incident light.
The maximum kinetic energy of photoelectrons is given as:
\[K{E_{\max }} = E - {W_0}\]
Where E is the energy of the photon and \[{W_0}\] is the work function.

Complete step by step solution:
Given Work function, \[{W_0} = 4.2{\rm{ eV}}\]
Wavelength, \[\lambda = 2000\mathop A\limits^o \]
We know the mass of an electron, m=\[9.1 \times {10^{ - 31}}kg\]
As we know that the energy of the photon is,
\[E = h\upsilon \]
Also written as
\[E = \dfrac{{hc}}{\lambda }\\
\Rightarrow E= \dfrac{{12375}}{{2000}}\\
\Rightarrow E= 6.18 eV\]
The maximum kinetic energy of photoelectrons is,
\[K{E_{\max }} = E - {W_0}\]
\[\Rightarrow \dfrac{1}{2}m{v_{\max }}^2 = E - {W_0}\]
\[\begin{array}{l}\dfrac{1}{2}m{v_{\max }}^2 = 6.18 - 4.2\\ \Rightarrow \dfrac{1}{2}m{v_{\max }}^2 {\rm{ = 1}}{\rm{.98 eV}}\\ \Rightarrow \dfrac{1}{2}m{v_{\max }}^2 {\rm{ = 1}}{\rm{.98}} \times {\rm{1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 19}}\\\therefore {\rm{ }}{{\rm{v}}_{\max }} = \sqrt {\dfrac{{{\rm{1}}{\rm{.98}} \times 2 \times {\rm{1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}} \\{\rm{ }} \approx {\rm{8}}{\rm{.4}} \times {\rm{1}}{{\rm{0}}^5}{\rm{ m/sec}}\\{\rm{ }}\end{array}\]
Therefore, the velocity of the fastest electron ejected from the surface will be \[8.4 \times {10^5}{\rm{ m/sec}}\].

Hence option A is the correct answer

Note: The photoelectric is a phenomenon where electrons are ejected from a metal surface when light of sufficient frequency is incident on it. Einstein suggested that light behaved like a particle and that each particle of light has energy called a photon. When a photon falls on the metal surface, the photon energy is transferred to the electron.