If the frequency of an emitted x-ray is 1.6 x 10$^{16}$Hz, the accelerating potential is?  [e = 1.6 x 10$^{-19}$J, h = 6.63 x 10-$^{34}$Js]

1. Analyze the Question

We are given the frequency ($f$) of an emitted X-ray and asked to find the accelerating potential ($V$) required to produce it. We are provided with the elementary charge ($e$) and Planck's constant ($h$).

2. Explain the Concept

In X-ray production, the kinetic energy gained by an electron accelerated through a potential difference $V$ is converted into the energy of the emitted X-ray photon. This is based on the principle of conservation of energy: Electrical Energy = Photon Energy

Where:

• $e$ = charge of an electron ($1.6 \times 10^{-19} \, \text{C}$)
• $V$ = accelerating potential (in Volts)
• $h$ = Planck's constant ($6.63 \times 10^{-34} \, \text{Js}$)
• $f$ = frequency of the X-ray ($1.6 \times 10^{16} \, \text{Hz}$)

3. Show the Solution

To find the potential $V$, we rearrange the formula:

Step 1: Substitute the given values into the equation:

Step 2: Simplify the expression: Notice that $1.6$ appears in both the numerator and the denominator, so they cancel out:

Step 3: Solve the powers of 10: Using the laws of indices ($10^a \times 10^b = 10^{a+b}$ and $\frac{10^a}{10^b} = 10^{a-b}$):

4. Identify the Answer

Comparing our result to the given options, we find that 66.3V matches option C.

5. Exam Tip

When you see a calculation involving $1.6 \times 10^{-19}$ (the charge of an electron) and $1.6 \times 10^{16}$ (or similar values) in a physics problem, always look for opportunities to cancel out the $1.6$ before doing any heavy multiplication. It saves time and reduces the chance of making a decimal point error!