Question
The wavelength $\lambda$ associated with a moving particle depends upon its mass m , its velocity v and Planck's constant h. Show dimensional relation between them.
✓
Answer
Suppose wavelength $\lambda$ associated with a moving particle depends upon (i) its mass (m), (ii) its velocity (v) and (iii) Planck's constant (h), then
$
\lambda=k m^a v^b h^c . .(1)
$
where, k is a dimensionless constant.
Representing the above equation in terms of its dimensions, we get
$
\begin{aligned}
& {\left[M^0 L^1 T^0\right]=[M]^a\left[LT^{-1}\right]^{b}\left[ML^2 T^{-1}\right]^{c}} \\
& \Rightarrow\left[M^0 L^1 T^0\right]=M^{a+c} L^{b+2 c} T^{-b-c} . .(2)
\end{aligned}
$
Comparing power of $M , L$ and T on both sides of equation (2), we get
$
a+c=0, b+2 c=1,-b-c=0
$
we get $a=-1, b=-1, c=+1$
putting the value of $a, b$, and $c$ in equation (1), we get
$
\begin{aligned}
& \lambda=k m^{-1} v^{-1} h^1 \\
& \lambda=\frac{k h}{m v}
\end{aligned}
$
Hence, the relation becomes $\lambda=\frac{k h}{m v}$ and it gives the de broglie wavelength of a moving particle.
$
\lambda=k m^a v^b h^c . .(1)
$
where, k is a dimensionless constant.
Representing the above equation in terms of its dimensions, we get
$
\begin{aligned}
& {\left[M^0 L^1 T^0\right]=[M]^a\left[LT^{-1}\right]^{b}\left[ML^2 T^{-1}\right]^{c}} \\
& \Rightarrow\left[M^0 L^1 T^0\right]=M^{a+c} L^{b+2 c} T^{-b-c} . .(2)
\end{aligned}
$
Comparing power of $M , L$ and T on both sides of equation (2), we get
$
a+c=0, b+2 c=1,-b-c=0
$
we get $a=-1, b=-1, c=+1$
putting the value of $a, b$, and $c$ in equation (1), we get
$
\begin{aligned}
& \lambda=k m^{-1} v^{-1} h^1 \\
& \lambda=\frac{k h}{m v}
\end{aligned}
$
Hence, the relation becomes $\lambda=\frac{k h}{m v}$ and it gives the de broglie wavelength of a moving particle.
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