We know, for a formula to be correct only when dimensions of LHS and dimensions of RHS must be equal.
E=⇒=.
⇒ Both LHS and RHS of Eq. (i), are dimensionless. Hence, the formula is dimensionally correct.
- In
E=E0e−, clearly the exponent is dimensionless because,
L and
L0 will have same dimension and dimensions of
E and
E0 will be equal, since both
E and
E0 will have dimensions of energy. Thus, this formula is dimensionally correct.
- In
E=Le−L∕E0, clearly the power of exponential
will have dimensions and hence the power of. exponent is not dimensionless, hence this equation is dimensionally incorrect.
- In
E=2()×e−L∕L0, here
will have different dimensions from
E, hence this formula is also dimensionally incorrect.