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

To calculate the standard deviation of the z-component of spin measurements from a Stern-Gerlach experiment, use the formula for standard deviation in a binomial distribution. With 46 particles showing spin up and 54 spin down, the standard deviation is found to be approximately 4.984.

Step-by-step explanation:

The question involves calculating the standard deviation of the z-component of spin measurements in a Stern-Gerlach experiment. Given that 46 particles yielded a spin of +½ℏ and 54 particles yielded a spin of -½ℏ, we can use these values to compute the standard deviation. The formula for the standard deviation σ in this binomial distribution is σ = √(np(1-p)), where n is the total number of trials and p is the probability of success (getting a +½ℏ spin result).

• Number of trials, n = 100
• Number of successes (spin up), k = 46
• Probability of success, p = k/n = 46/100

Using these values, the standard deviation is:

σ = √(100 * (46/100) * (1 - 46/100))
σ = √(100 * 0.46 * 0.54)
σ = √(24.84)
σ = 4.984

The standard deviation of the z-component of spin measurements in this experiment is approximately 4.984.

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Answer 2

The standard deviation is 0.4984
`\hbar`

Step-by-step explanation:

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

`\sigma=\sqrt{(1)/(n) \sum_(i=1)^(100) (\mu-x_i)^2`

Here μ is the mean which is calculated as follows

`\mu=(\sum_(i=1)^(100) x_i)/(n)\\\mu=(46* (\hbar)/(2)+54* (-\hbar)/(2))/(100)\\\mu=(-4 \hbar)/(100)\\\mu=-0.04 \hbar`

Now the standard deviation is given as

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

So the standard deviation is 0.4984
`\hbar`