Question

A photon has energy \(6\text{ eV}\). Its frequency is approximately:

• A. \(1.45\times10^{15}\text{ Hz}\)
• B. \(6\times10^{15}\text{ Hz}\)
• C. \(9\times10^{14}\text{ Hz}\)
• D. \(3\times10^{8}\text{ Hz}\)

Hint:

Remember: \[ 1\text{ eV}=1.6\times10^{-19}\text{ J} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Planck's quantum theory, electromagnetic radiation travels as discrete packets of energy called photons. The energy carried by a single photon scales linearly with its oscillation frequency. When given energy in non-SI units like electron-volts ($\text{eV}$), we must convert it to Joules ($\text{J}$) to maintain consistency with standard metric constants.
Step 2: Key Formula or Approach:
The Planck-Einstein equation relates photon energy ($E$) to frequency ($\nu$): $$ E = h\nu \implies \nu = \frac{E}{h} $$ Let's specify the required physical conversion constants: - Planck's constant ($h$) $\approx 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$ - Conversion factor: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules}$
Step 3: Detailed Explanation:
Let's convert the given energy value from electron-volts into Joules: $$ E = 6 \text{ eV} = 6 \times 1.6 \times 10^{-19} \text{ J} $$ $$ E = 9.6 \times 10^{-19} \text{ J} $$ Now, substitute this energy value and Planck's constant into the frequency formula: $$ \nu = \frac{9.6 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J}\cdot\text{s}} $$ Isolate the base numbers and powers of 10 to simplify the calculation: $$ \nu = \left( \frac{9.6}{6.63} \right) \times 10^{-19 - (-34)} \text{ Hz} $$ $$ \nu \approx 1.4479 \times 10^{15} \text{ Hz} $$ Rounding this value to two decimal places gives approximately $1.45 \times 10^{15} \text{ Hz}$. This matches option (A).
Step 4: Final Answer:
The frequency of the photon is approximately $1.45 \times 10^{15}$ Hz.