JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 23)
The dimension of stopping potential V0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :
h1/3 G2/3 c1/3 A–1
h0 c5 G-1 A-1
h2/3 c5/3 G1/3 A–1
h2 G3/2 c1/3 A–1
Explanation
V0 $$ \propto $$ hPcQGRIS
[V0] = [M1L2T–3A–1]
[c] = [L1T–1]
[h] = [M1L2T–1]
[G] = [M–1L3T–2]
[I] = [A]
$$ \therefore $$ [M1L2T–3A–1] = [MP–R L2P+Q+3R T–P–Q–2R AS]
Comparing dimension of M, L, T, A, we get
P – R = 1 ; 2P + Q + 3R = 2
– P – Q – 2R = – 3 ; S = – 1
$$ \Rightarrow $$ P = 0, Q = 5, R = –1, S = –1
$$ \therefore $$ V0 $$ \propto $$ h0 c5 G-1 A-1
[V0] = [M1L2T–3A–1]
[c] = [L1T–1]
[h] = [M1L2T–1]
[G] = [M–1L3T–2]
[I] = [A]
$$ \therefore $$ [M1L2T–3A–1] = [MP–R L2P+Q+3R T–P–Q–2R AS]
Comparing dimension of M, L, T, A, we get
P – R = 1 ; 2P + Q + 3R = 2
– P – Q – 2R = – 3 ; S = – 1
$$ \Rightarrow $$ P = 0, Q = 5, R = –1, S = –1
$$ \therefore $$ V0 $$ \propto $$ h0 c5 G-1 A-1
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