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Question: The dimensions of stopping potential V in photoelectric effect in units of Planck's constant “h” ,sp...

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−23c−13G43A−1{h^{\dfrac{{ - 2}}{3}}}{c^{\dfrac{{ - 1}}{3}}}{G^{\dfrac{4}{3}}}{A^{ - 1}}
(B) h2c13G32A−1{h^2}{c^{\dfrac{1}{3}}}{G^{\dfrac{3}{2}}}{A^{ - 1}}
(C) h0c5G−1A−1{h^0}{c^5}{G^{ - 1}}{A^{ - 1}}
(D) h13c13G23A−1{h^{\dfrac{1}{3}}}{c^{\dfrac{1}{3}}}{G^{\dfrac{2}{3}}}{A^{ - 1}}

Explanation

Solution

Hint The dimensional formulas of the physical quantities are substituted in the known relation V=haIbGccdV = {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= [ML2T−3A−1][M{L^2}{T^{ - 3}}{A^{ - 1}}]
h= [ML2T−1][M{L^2}{T^{ - 1}}]
c= [M0L1T−1][{M^0}{L^1}{T^{ - 1}}]
G= [M−1L3T−2][{M^{ - 1}}{L^3}{T^{ - 2}}]
I= [A]
Writing stopping potential in terms of the required quantities,
V=haIbGccdV = {h^a}{I^b}{G^c}{c^d}
[ML2T−1]a[Ab][M−1L3T−2]c[M0L2T−1]d{[M{L^2}{T^{ - 1}}]^a}[{A^b}]{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}{[{M^0}{L^2}{T^{ - 1}}]^d}

[ML2T−3A−1]=[M]a−c[L]2a+3c+d[T]−a−2c−d[A]b   [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  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=0a = 0
b=−1b = - 1
c=−1c = - 1
d=1d = 1
Substituting this value in the initial equation, we get
V=h0c5G−1A−1V = {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.