Question

Let ${{\text{k}}_1}$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength ${\lambda _1}$ and ${{\text{k}}_2}$ responding to ${\lambda _2}$, where ${\lambda _1} = 2{\lambda _2}$, then:
A) $2{{\text{k}}_1} = {{\text{k}}_2}$
B) ${{\text{k}}_{\text{1}}}{\text{ = 2}}{{\text{k}}_{\text{2}}}$
C) ${{\text{k}}_{\text{1}}}{\text{ < 2}}{{\text{k}}_{\text{2}}}$
D) ${{\text{k}}_{\text{2}}} > 2{{\text{k}}_{\text{1}}}$

Answer 1

Hint: By the action of light, metal atoms emit electrons. If one is specific, then metal atoms emit electrons from the surface when light is incident on the surface. The phenomenon is termed the photoelectric effect, and the emitted electrons are called photoelectrons. The energy of these photoelectrons is significantly related to the property of the incident light.

Complete step by step solution:
Light Waves carry energy in the form of quanta or photons. When the light is incident on a photosensitive metal surface, metals emit electrons from its surface. The photons in the incident wave interact with the far most electron of an atom, resulting in emitted electrons from the metal surface. The metal surface emits electrons till the light is incident on the surface.
We use Einstein's Photoelectric Equation-
$k = h\nu - W$…………….$(1)$
Here, $k$ = maximum kinetic energy of the photoelectrons
$h$ is the Planck's constant
$\nu $ is the frequency of the incident light
$W$ is the threshold energy, i.e., minimum work required to release photoelectrons
In $(1)$, we put $\nu = \dfrac{c}{\lambda }$ and get-
$k = \dfrac{{hc}}{\lambda } - W$…………….$(2)$
where $\lambda $ is the wavelength of the incident light and
$c$ is the speed of light.
Now using equation $(2)$, we have-
$\Rightarrow {k_1} = \dfrac{{hc}}{{{\lambda _1}}} - W$
$ \Rightarrow {k_1} = \dfrac{{hc}}{{2{\lambda _2}}} - W$
$ \Rightarrow \dfrac{{hc}}{{2{\lambda _2}}} = W + {k_1}$…………$(3)$
Also,
$\Rightarrow {k_2} = \dfrac{{hc}}{{{\lambda _2}}} - W$
$\Rightarrow \dfrac{{hc}}{{{\lambda _2}}} = W + {k_2}$…………$(4)$
We divide the equation $(4)$ by $(2)$ and then compare with the equation $(3)$-
$\Rightarrow W + {k_1} = \dfrac{W}{2} + \dfrac{{{k_2}}}{2}$
$ \Rightarrow 2W + 2{k_1} = W + {k_2}$
$ \Rightarrow {k_2} = 2{k_1} + W$
Hence, we have ${k_2} > 2{k_1}$

Therefore, the required answer is ${k_2} > 2{k_1}$, and option (D) is the correct answer.

Note: It is easier to emit electrons from atoms with lower threshold energy. The maximum kinetic energy of the photoelectrons is directly proportional to the frequency of the incident light. Again, the photoelectrons' maximum kinetic energy is inversely proportional to the wavelength of the incident light. Between a red light and blue light of the same intensity, the latter has more energy.