The dimension of stopping potential $V_{0}$ in photoelectric effect in units of Planck's constant $h$, speed of light $c$, Gravitational constant $G$, and ampere $A$ is:

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Aicharya · Accepted Answer

V = K ( h ) a ( I ) b ( G ) c ( C ) d (V is voltage) We know [ h ] = M L 2 T - 1 [ I ] = A [ G ] = M - 1 L 3 T - 2 [ C ] = L T - 1 [ V ] = M L 2 T - 3 A - 1 M L 2 T - 3 A - 1 = ( M L 2 T - 1 ) a ( A ) b ( M - 1 L 3 T - 2 ) c ( L T - 1 ) d M L 2 T - 3 A - 1 = M a - c L 2 a + 3 c + d T - a - 2 c - d A b a - c = 1 &#8201; . . . . . ( 1 ) 2 a + 3 c + d = 2 &#8201; . . . . . ( 2 ) - a - 2 c - d = - 3 &#8201; . . . . . ( 3 ) b = - 1 &#8201; &#8201; . . . ( 4 ) On solving c = - 1 a = 0 d = 5 , &#8201; b = - 1 V = K ( h ) 0 ( A ) - 1 ( G ) - 1 ( C ) 5

The dimension of stopping potential $V_{0}$ in photoelectric effect in units of Planck's constant $h$, speed…

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