# Easy question calculating uncertainty

LBRRIT2390

## Homework Statement

What is the mass uncertainty, $$\delta$$m = m+ - m, of the particle?

## Homework Equations

m = $$\frac{R*|q|*B}{v}$$

where
R = (6.4$$\pm$$0.1) cm (convert to meters so 0.064$$\pm$$0.001)
|q| = 1.6*10-19
B = (2.50$$\pm0.02$$) T
v = (5.25$$\pm0.04$$)x107

## The Attempt at a Solution

I figured out that m = m = 4.88*10-28kg

When I tried to find the upper limit I got 4.95x10-28
and the lower limit, 4.80x10-28

My professor's answer sheet says that m+ = 5.03 x 10-28 kg
and $$\delta$$m = 0.2x10-28 kg.

First of all, I don't know how he determined that $$\delta$$m = 0.2x10-28 kg. Can someone help me?

Second of all, if m = (4.88*10-28kg) and $$\delta$$m = 0.2x10-28 kg then shouldn't m+ = 5.08 x 10-28 kg?? NOT 5.03 x 10-28 kg??

Homework Helper
Technically, the correct formula from error analysis is

$$\delta f(x_1, \cdots, x_n) = \sqrt{ \sum_{i = 1}^n \left( \frac{\partial f(x_1, \cdots, x_n)}{\partial x_i} \delta x_i \right)^2 }$$
where $\delta x_i$ is the uncertainty in xi.

So in this particular case, you would get
$$(\delta m)^2 = \left( \frac{\partial}{\partial R} \frac{R|q|B}{v} \right)^2 (\delta R)^2 + \left( \frac{\partial}{\partial |q|} \frac{R|q|B}{v} \right)^2 (\delta |q|)^2 + \left( \frac{\partial}{\partial B} \frac{R|q|B}{v} \right)^2 (\delta B)^2 + \left( \frac{\partial}{\partial v} \frac{R|q|B}{v} \right)^2 (\delta v)^2.$$