Question

When a light of frequency ${v_1}$ is incident on a metal surface, the photoelectrons emitted have twice the kinetic energy as did the photoelectrons emitted when the same metal has irradiated with light of frequency ${v_2}$. What will be the value of threshold frequency?
A. ${{\text{v}}_0} = {{\text{v}}_1} - {{\text{v}}_2}$
B. ${{\text{v}}_0} = {{\text{v}}_1} - 2{{\text{v}}_2}$
C. ${{\text{v}}_0} = 2{{\text{v}}_2} - {{\text{v}}_1}$
D. ${{\text{v}}_0} = {{\text{v}}_1} + {{\text{v}}_2}$

Answer 1

Photoelectrons are the electrons that are formed when an energetic photon of light hits a molecule, and the photoelectron spectrum is determined by analysing the spectrum of energies that they contain. The findings contradict traditional electromagnetism, which states that continuous light waves transmit energy to electrons, which are subsequently released after they have accumulated enough energy. Here we use the Kinetic energy of photoelectrons concept to solve the answer.

Complete step by step answer:
Planck's innovative concept that light is a particle was used by Einstein to overcome this difficulty. The energy carried by each particle of light (known as a quanta or photon) is determined by the frequency (v) of the light, as shown:
$E = hv$
Where $h$ is the Planck constant, which is $6.6261{\text{ }} \times {10^{ - 34}}$ Js.

Because light is packaged into photons, Einstein hypothesised that when a photon collides with a metal's surface, the full photon's energy is transmitted to the electron. A portion of this energy is utilised to free the electron from the hold of the metal atom, with the remainder going to the expelled electron as kinetic energy.

During the impact, electrons released from underneath the metal surface lose some kinetic energy. Surface electrons, on the other hand, carry all of the proton's kinetic energy and have the highest kinetic energy. This may be expressed numerically as: Photon energy is equal to the energy necessary to expel an electron (work function) plus the electron's maximum kinetic energy.
W + KE = E \\\ \Rightarrow W + KE = hv \\\ \Rightarrow KE = hv – w \\\
${v_o}$ electrons are simply expelled at the threshold frequency and have no kinetic energy.There is no electron emission below this frequency. As a result, the work function of the metal must be the energy of a photon with this frequency.
${\text{h}}{{\text{v}}_1} = {\text{h}}{{\text{v}}_0} + {\mathbf{K}}{{\mathbf{E}}_1}$
$\Rightarrow {\text{h}}{{\text{v}}_2} = {\text{h}}{{\text{v}}_0} + {\mathbf{K}}{{\mathbf{E}}_2}$
But according to given we can write
${\text{K}}{{\text{E}}_{\text{1}}}{\text{ = 2K}}{{\text{E}}_{\text{2}}}$
Upon substituting we get,
${\text{h}}{{\text{v}}_1} = {\text{h}}{{\text{v}}_0} + 2{\mathbf{K}}{{\mathbf{E}}_2}$

From the above equations one can deduce
${\text{h}}{{\text{v}}_2}{\text{ - h}}{{\text{v}}_0} = {\mathbf{K}}{{\mathbf{E}}_2}$
Hence upon all calculations we get,
${\text{h}}{{\text{v}}_1} = {\text{h}}{{\text{v}}_0} + 2\left( {\;{\text{h}}{{\text{v}}_2} - {\text{h}}{{\text{v}}_0}} \right)$
$\Rightarrow {\text{h}}{{\text{v}}_1} = {\text{h}}{{\text{v}}_0} + 2\;{\text{h}}{{\text{v}}_2} - 2\;{\text{h}}{{\text{v}}_0}$
${\text{h}}{{\text{v}}_1} = 2{\text{h}}{{\text{v}}_2} - {\text{h}}{{\text{v}}_0}$
$\Rightarrow {\text{h}}{{\text{v}}_0} = 2{\text{h}}{{\text{v}}_2} - {\text{h}}{{\text{v}}_1}$
$\therefore {{\text{v}}_0} = 2{{\text{v}}_2} - {{\text{v}}_1}$
So the threshold energy is $\Rightarrow {{\text{v}}_0} = 2{{\text{v}}_2} - {{\text{v}}_1}$

Hence, option C is correct.

Note: Make no error while substituting for Einstein photoelectric equation. J J Thomson was the first to discover the electron. The minimal frequency necessary for the photoelectric effect to occur is known as the threshold frequency. Typically, alkali metal is utilised to demonstrate photoelectric effect. Copper and sodium are two examples of metals that show the photoelectric effect. The photoelectric effect is caused by ultraviolet light, which is a form of light. Electric eyes in door openers, light metres in photography, photo static copying, and solar panels are all examples of photoelectric effect uses.