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An electron beam under an acceleration voltage is incident on a single crystal with a simple cubic structure, whose lattice constant is a=3.359 A . The incident electron beam is along the [ 1 1 1 } direction, and diffracted by the ( 1 1 0 ) planes in the first order. Find out the angle in degrees between the incident electron beam and the ( 1 1 0 ) planes.

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The first step in solving this problem is to define the direction of the incident electron beam and the planes from which the beam is diffracted. Here, the incident electron beam direction, denoted as 'd', can be represented as a vector [1, 1, 1]. The planes from which the beam is diffracted are represented by vector 'p' or [1, 1, 0].

Normally, to calculate the metrics related to these vectors such as the dot product and norms. The dot product of two vectors gives a scalar value representing the multiplication of the corresponding entries.

Next, calculate the dot product of vectors 'd' and 'p' using the standard formula, which sums the products of the corresponding entries of the two sequences of numbers. By executing the formula, we received the dot product as 2.

To find the norm (or length) of a vector, square each component of the vector, sum these squares, and finally, take the square root of this sum. Executing these steps, we get the norms of 'd' and 'p' which are 1.732 and 1.414 respectively.

Now, to find the angle between the two vectors 'd' and 'p', we use the formula: cos(θ) = (d . p) / (||d|| ||p||), where 'd . p' is the dot product and ||d|| and ||p|| are the norms calculated in previous steps. Substituting in the values from earlier steps, cos(θ) = 2/ (1.732 * 1.414) = 0.816.

The above value (0.816) is the measure of the cosine of our angle in radians. To convert this to the angle itself, we use the arccosine operation, which is the inverse of cosine, resulting in a value of 0.615 radians.

Finally, convert this angle from radians to degrees by multiplying by (180/π), getting a result of 35.264 degrees.

So, the angle between the electron beam and the (1, 1, 0) planes is 35.264 degrees.

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