X rays with a wavelength of 71 pm are directed onto a gold foil and eject tightly bound electrons from the gold atoms. The ejected electrons then move in circular paths of radius r in a region of uniform magnetic field `vecB`. For the fastest of the ejected electrons, the product Br is equal to `1.88 xx 10^(-4)` T.m. Find (a) the maximum kinetic energy of those electrons and (b) the work done in removing them from the gold atoms.

Explanation

To solve the problem, we need to find two things: (a) the maximum kinetic energy of the ejected electrons and (b) the work done in removing them from the gold atoms.

### Step-by-step Solution

**(a) Maximum Kinetic Energy of the Ejected Electrons**

1. **Identify the given values:**
- Wavelength of X-rays, \( \lambda = 71 \, \text{pm} = 71 \times 10^{-12} \, \text{m} \)
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