The dimension of stopping potential V0 in photoelectric effect in units of Planck’s constant ‘h’, speed of light ‘C’ and universal gravitational constant ‘G’ and ampere A is:
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a
h1/3G2/3,C1/3A−1
b
h0 , C5, G-1, A−1
c
h−2/3,C−1/3,G4/3,A−1
d
h2 , G3/2, C1/3 , A−1
answer is B.
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Detailed Solution
V=KhaAbGcCd (V is voltage)We know h=ML2T−1[A] = A[G] = M-1L3T-2[C] = LT-1[V] = ML2T-3A−1 comparing dimensions on both sides, ML2T-3A−1 = ML2T−1aAbM−1L3T−2cLT−1d ⇒ML2T-3A−1 = Ma−cL2a+3c+dT−a−2c−dAbComparing the powers, a−c=1..................(1)b=−1.................2−a−2c−d = −3 .........32a+3c+d = 2 .............4Add 3 and 4 , a+c = −1 .........5Add 1 and 5 , 2a =0⇒a=0∴c=−1Put values of a and c in 3,2−d=−3⇒d=5Thus, V=K h0A−1G−1C5