Question

An ensemble of 100 identical particles is sent through a Stern-Gerlach apparatus and the z-component of spin is measured. 46 yield the value +\frac{\hbar}{2}+ ℏ 2 while the other 54 give -\frac{\hbar}{2}− ℏ 2. Compute the standard deviation of the measurements.

Answer 1

Answer:

The standard deviation is 0.4984 $\hbar$

Step-by-step explanation:

In order to find standard deviation, The equation is given as

$\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{100} (\mu-x_i)^2$

Here μ is the mean which is calculated as follows

                                       $\mu=\frac{\sum_{i=1}^{100} x_i}{n}\\\mu=\frac{46\times \frac{\hbar}{2}+54\times \frac{-\hbar}{2}}{100}\\\mu=\frac{-4 \hbar}{100}\\\mu=-0.04 \hbar$

Now the standard deviation is given as

                      $\sigma=\sqrt{\frac{1}{100} \sum_{i=1}^{100} (-0.04 \hbar-x_i)^2}\\\sigma=\sqrt{\frac{1}{100} [[46 \times(-0.04 \hbar-0.5 \hbar)^2]+[54 \times(-0.04 \hbar+0.5 \hbar)^2]}]\\\sigma=\sqrt{\frac{1}{100} [[46 \times(-0.54 \hbar)^2]+[54 \times(0.46 \hbar)^2]}]\\\sigma=\sqrt{\frac{1}{100} [[46 \times(0.2916 \hbar)]+[54 \times(0.2116 \hbar)]}]\\\sigma=\sqrt{\frac{1}{100} [13.4136 \hbar+11.4264 \hbar}]\\\sigma=\sqrt{\frac{24.84 \hbar}{100}}\\\sigma =0.4984 \hbar$

So the standard deviation is 0.4984 $\hbar$