Question

The ratio of slopes of ${{\text{K}}_{{\text{max}}}}$ vs $\upsilon $ and ${\upsilon _0}$ vs $\upsilon $ curves in the photoelectric effects gives:
[$\upsilon = $ frequency, ${{\text{K}}_{{\text{max}}}} = $ maximum kinetic energy, ${\upsilon _0} = $ stopping potential]
A. the ratio of Planck’s constant of electronic charge
B. work function
C. Planck’s constant
D. charge of electron

Answer 1

Hint:When light of suitable frequency is incident on a metal surface then electrons are emitted from the surface called as photoelectric effect. Stopping potential is applied to stop this ejection of electrons. It is an external potential difference.

Complete step by step answer:
The maximum kinetic energy can be determined from the stopping potential by using conservation of energy. Thus loss of kinetic energy is equal to the gain of kinetic energy. Thus it can be expressed as:
${{\text{K}}_{{\text{max}}}} = {\text{e}}{\upsilon _0}$, where ${{\text{K}}_{{\text{max}}}}$ is the maximum kinetic energy, ${\text{e}}$ is the charge of electron $\left( {1.6 \times {{10}^{ - 19}}{\text{C}}} \right)$ and ${\upsilon _0}$ is the stopping potential.
Also maximum kinetic energy of a photoelectron is expressed as:
${{\text{K}}_{\max }} = {\text{h}}\upsilon - {\text{h}}{\upsilon _0}$, where ${\text{h}}$ is the Planck’s constant $\left( {6.636 \times {{10}^{ - 34}}{\text{J}}.{\text{s}}} \right)$, ${\upsilon _0}$ is the stopping potential and $\upsilon $ is the frequency.
${\text{h}}{\upsilon _0}$ is the work function of the metal which is the minimum energy of radiation to eject an electron from the metal surface.
Combining both equations, we get
${\text{e}}{\upsilon _0} = {\text{h}}\upsilon - {\text{h}}{\upsilon _0}$
From the above equation, stopping potential can be expressed as:
${\upsilon _0} = \dfrac{{\text{h}}}{{\text{e}}}\upsilon - \dfrac{{\text{h}}}{{\text{e}}}{\upsilon _0}$
We can compare the above equation with the straight line equation ${\text{y}} = {\text{mx}} + {\text{c}}$, where ${\text{m}}$ is the slope.
Thus in ${\upsilon _0} = \dfrac{{\text{h}}}{{\text{e}}}\upsilon - \dfrac{{\text{h}}}{{\text{e}}}{\upsilon _0}$, ${\text{y}}$ is ${\upsilon _0}$, ${\text{m}}$ is $\dfrac{{\text{h}}}{{\text{e}}}$, ${\text{x}}$ is $\upsilon $ and ${\text{c}}$ is $ - \dfrac{{\text{h}}}{{\text{e}}}{\upsilon _0}$.
In ${\upsilon _0} = \dfrac{{\text{h}}}{{\text{e}}}\upsilon - \dfrac{{\text{h}}}{{\text{e}}}{\upsilon _0}$, the slope of ${\upsilon _0}$ vs $\upsilon $ curves, ${{\text{m}}_1}$ is $\dfrac{{\text{h}}}{{\text{e}}}$.
And, the slope of ${{\text{K}}_{{\text{max}}}}$ vs $\upsilon $ curves, Planck’s constant, ${\text{h}}$ is the slope.
Thus ${{\text{m}}_2} = {\text{h}}$
When both slopes are combined, the ratio of slopes of ${{\text{K}}_{{\text{max}}}}$ vs $\upsilon $ and ${\upsilon _0}$ vs $\upsilon $ curves in the photoelectric effects $ = \dfrac{{{{\text{m}}_2}}}{{{{\text{m}}_1}}} = \dfrac{{\text{h}}}{{\dfrac{{\text{h}}}{{\text{e}}}}} = {\text{h}} \times \dfrac{{\text{e}}}{{\text{h}}} = {\text{e}}$
Thus the ratio of slopes is equal to the charge of electron, ${\text{e = 1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 19}}{\text{C}}$

Hence, the correct option is D.

Note:
All photons do not have the same amount of energy. Some photons collide with other particles and transfer their energy. And the maximum theoretical kinetic energy is ${{\text{K}}_{\max }} = {\text{h}}\upsilon - {\text{h}}{\upsilon _0}$. Different characteristic properties of different substances can also be determined from the graphs.