Question

$\mathtt{\large{ \red{\underline{\underline{question : }}}}} \\ \\ \bigstar \sf \: find \: the \: dimension \: of \: \alpha \\ \\ \displaystyle \sf \longrightarrow \: \alpha \: in \: p = p0 {e}^{ - \alpha {t}^{2} } \\ \\ \displaystyle \sf \longrightarrow \: where \: t = time \: and \: p0 \: = p \: not$
Koi toh solve kr do ​

Answer 1

Answer:

$\boxed{\mathfrak{Dimension \ formula \ of \ alpha = [T^{-2}]}}$

Explanation:

Given:

$\rm p = p_o e^{-\alpha t^2}$

t → Time

Dimensional formula of time = [T]

As we know exponential functions are dimensionless.

$\therefore \rm \alpha t^2$ is dimensionless i.e.

$\rm \implies [\alpha T^2] = [M^0L^0T^0] \\ \\ \rm \implies [\alpha ] = \frac{ [M^0L^0T^0]}{ [T^2]} \\ \\ \rm \implies [\alpha ] = [M^0L^0T^ {-2}] \\ \\ \rm \implies [\alpha ] = [T^ {-2}]$

Answer 2

Dimension of time = [T]

we know that exponential functions are dimensionless

so, αt² = dimensionless = [M⁰L⁰T⁰]

[αT²] = [M⁰L⁰T⁰]

α = [M⁰L⁰T⁰]/T²

α = [M⁰L⁰T`²]

α = [T^-2]