
The dimensions of stopping potential V in photoelectric effect in units of Planck's constant “h” ,speed of light “c” and Gravitational constant ‘G” and Ampere “A” is
(A) \[{h^{\dfrac{{ - 2}}{3}}}{c^{\dfrac{{ - 1}}{3}}}{G^{\dfrac{4}{3}}}{A^{ - 1}}\]
(B) \[{h^2}{c^{\dfrac{1}{3}}}{G^{\dfrac{3}{2}}}{A^{ - 1}}\]
(C) \[{h^0}{c^5}{G^{ - 1}}{A^{ - 1}}\]
(D) \[{h^{\dfrac{1}{3}}}{c^{\dfrac{1}{3}}}{G^{\dfrac{2}{3}}}{A^{ - 1}}\]
Answer
219k+ views
Hint The dimensional formulas of the physical quantities are substituted in the known relation \[V = {h^a}{I^b}{G^c}{c^d}\] . This equated to the dimensions of stopping potential. After multiplication of the powers the corresponding powers of mass, length, time and ampere are equated to get equations in terms of the powers a, b, c and d. Solve the equations simultaneously to find the values of a, b, c and d. Substitute them in the initial equation to get the dimensions of stopping potential.
Complete step-by-step solution
Expressing each quantity in terms of the fundamental units,
V= \[[M{L^2}{T^{ - 3}}{A^{ - 1}}]\]
h= \[[M{L^2}{T^{ - 1}}]\]
c= \[[{M^0}{L^1}{T^{ - 1}}]\]
G= \[[{M^{ - 1}}{L^3}{T^{ - 2}}]\]
I= [A]
Writing stopping potential in terms of the required quantities,
\[V = {h^a}{I^b}{G^c}{c^d}\]
\[{[M{L^2}{T^{ - 1}}]^a}[{A^b}]{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}{[{M^0}{L^2}{T^{ - 1}}]^d}\]
\[
[M{L^2}{T^{ - 3}}{A^{ - 1}}] = {[M]^{a - c}}{[L]^{2a + 3c + d}}{[T]^{ - a - 2c - d}}{[A]^b} \\
\\
\]
Equating both sides to find the required form,
\[
1 = a - c \\
2 = 2a + 3c + d \\
- 3 = - a - 2c - d \\
- a = b \\
\]
After solving these 4 linear equations in 4 variables, we get
\[a = 0\]
\[b = - 1\]
\[c = - 1\]
\[d = 1\]
Substituting this value in the initial equation, we get
\[V = {h^0}{c^5}{G^{ - 1}}{A^{ - 1}}\]
Therefore, the required answer is option C
Note
Dimensional analysis has other applications as well.
1. To find the units of physical quantities in a given system of units.
2. To find dimensions of physical constants or coefficients.
3. To convert physical quantities from one system to another.
4. To check dimensional correctness of a physical relation.
5. To derive new relations.
Complete step-by-step solution
Expressing each quantity in terms of the fundamental units,
V= \[[M{L^2}{T^{ - 3}}{A^{ - 1}}]\]
h= \[[M{L^2}{T^{ - 1}}]\]
c= \[[{M^0}{L^1}{T^{ - 1}}]\]
G= \[[{M^{ - 1}}{L^3}{T^{ - 2}}]\]
I= [A]
Writing stopping potential in terms of the required quantities,
\[V = {h^a}{I^b}{G^c}{c^d}\]
\[{[M{L^2}{T^{ - 1}}]^a}[{A^b}]{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}{[{M^0}{L^2}{T^{ - 1}}]^d}\]
\[
[M{L^2}{T^{ - 3}}{A^{ - 1}}] = {[M]^{a - c}}{[L]^{2a + 3c + d}}{[T]^{ - a - 2c - d}}{[A]^b} \\
\\
\]
Equating both sides to find the required form,
\[
1 = a - c \\
2 = 2a + 3c + d \\
- 3 = - a - 2c - d \\
- a = b \\
\]
After solving these 4 linear equations in 4 variables, we get
\[a = 0\]
\[b = - 1\]
\[c = - 1\]
\[d = 1\]
Substituting this value in the initial equation, we get
\[V = {h^0}{c^5}{G^{ - 1}}{A^{ - 1}}\]
Therefore, the required answer is option C
Note
Dimensional analysis has other applications as well.
1. To find the units of physical quantities in a given system of units.
2. To find dimensions of physical constants or coefficients.
3. To convert physical quantities from one system to another.
4. To check dimensional correctness of a physical relation.
5. To derive new relations.
Recently Updated Pages
A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

ill in the blanks Pure tungsten has A Low resistivity class 12 physics JEE_Main

The value of the resistor RS needed in the DC voltage class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

