The answer to one of the question is Δ λ = − λ m v 2 2Ψ cos θ . Here symbols have their usual notations. Ψ can represent

m v 2 2 ψ = ( λ Δ λ ⋅ c o s θ 1 ) 2 2Ψ = m v 2 ( constant ) dimension of Ψ = Energy = Dimension of mv 2

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The answer to one of the question is $Δ λ = - λ \frac{2Ψ}{m v ^{2}} cos θ$ . Here symbols have their usual notations. $Ψ$ can represent

Answer:

Solution:

$\frac{2 ψ}{m v ^{2}} = (\frac{Δ λ}{λ} \cdot \frac{1}{c o s θ})^{2} 2Ψ = m v^{2} ($ constant $)$

dimension of $Ψ =$ Energy $=$ Dimension of $mv^{2}$

Stream:JEESubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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