The given problem requires us to find the dimensional formula for Planck's constant \( h \). The de-Broglie wavelength formula given is:
\[\lambda = \frac{h}{\sqrt{2mE}}\]To find the dimensional formula of \( h \), we will utilize the given equation and analyze the dimensions involved. The de-Broglie wavelength formula can be rearranged as:
\(h = \lambda \times \sqrt{2mE}\)
Let's express the dimensions of each component:
Using these, we calculate the dimensions for \( \sqrt{2mE} \):
\(\sqrt{2mE} = \sqrt{[M][ML^2T^{-2}]} = \sqrt{[M^2L^2T^{-2}]}\)
Simplifying, we get:
\([MLT^{-1}]\)
Substituting back into the formula for \( h \):
\(h = \lambda \times \sqrt{2mE} = [L] \times [MLT^{-1}] = [ML^2T^{-1}]\)
Therefore, the dimensional formula for Planck's constant \( h \) is:
[ML2T-1]
This matches the provided correct answer option, which is:
\([ML^2T^{-1}]\)
To conclude, the correct dimensional formula for Planck's constant is indeed \([ML^2T^{-1}]\), and the selected option is correct. The other options do not match this dimensional analysis.
We are given the equation for the de-Broglie wavelength:
\[ \lambda = \frac{h}{\sqrt{2mE}}. \]
From this equation, rearranging to solve for \( h \):
\[ h = \lambda \sqrt{2mE}. \]
Now, let's find the dimensional formula for \( h \):
Now, substitute these into the equation:
\[ [h] = [L] \times \sqrt{[M] \times [ML^2T^{-2}]}. \]
Simplifying the terms inside the square root:
\[ [h] = [L] \times \sqrt{[M] \times [M][L^2][T^{-2}]} = [L] \times \sqrt{[M^2L^2T^{-2}]} = [L] \times [MLT^{-1}]. \]
Thus, the dimensional formula for \( h \) is:
\[ [h] = [ML^2T^{-1}]. \]
Therefore, the correct answer is Option (2).
If \( \lambda \) and \( K \) are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals: