The dimension of stopping potential V_0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :   
Option: 1 h^{-2/3}c^{-1/3}G^{4/3}A^{-1}
 
Option: 2 [h]^0[G]^{-1}[c]^5[A]^{-1}

Option: 3 h^{2}G^{3/2}c^{1/3}A^{-1}  

Option: 4 h^{1/3}G^{2/3}c^{1/3}A^{-1}
 

Answers (1)

 

 

 Let V_0=[h]^p[G]^q[c]^r[A]^s

Now, K_{max}=eV_0\Rightarrow [K_{max}]=[eV_0]\Rightarrow [V_0]=\frac{[K_{max}]}{[e]}\Rightarrow [V_0]=\frac{[ML^2T^{-2}]}{[AT]}\therefore [V_0]=[ML^2T^{-3}A^{-1}]

E=hf\Rightarrow [h]=\frac{[E]}{[f]}\Rightarrow \frac{[ML^2T^{-2}]}{[T^{-1}]}\Rightarrow [h]=[ML^2T^{-1}]

 

F=\frac{Gm^2}{r^2}\Rightarrow [G]=\frac{[F][r^2]}{[m^2]}=\frac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]

So, [ML^2T^{-3}A^{-1}]=[ML^2T^{-1}]^p[M^{-1}L^3T^{-2}]^q[LT^{-1}]^r[A]^s

From here we will get: p-q=1---------(1)

2p+3q+r=2----------(2)

-p-2q-r=-3----------(3)

s=-1-----------(4)

From equation 1, 2,3 and 4 we will get: p=0,q=-1, r=5 and s=-1

So, [V_0]=[h]^0[G]^{-1}[c]^5[A]^{-1}

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