Question

Light of wavelength \(400\ \text{nm}\) falls on a metal surface. The energy of one photon is approximately: \[ \left( h=6.63\times10^{-34}\ \text{J s},\quad c=3\times10^8\ \text{m/s} \right) \]

• A. \(4.97\times10^{-19}\ \text{J}\)
• B. \(2.5\times10^{-19}\ \text{J}\)
• C. \(1.2\times10^{-18}\ \text{J}\)
• D. \(6.63\times10^{-34}\ \text{J}\)

Hint:

Photon energy: \[ E=\frac{hc}{\lambda} \] Smaller wavelength means: \[ \text{higher photon energy} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Planck's Quantum Theory, electromagnetic radiation is not continuous but consists of discrete "packets" of energy called photons.
The energy of a single photon depends on the frequency of the radiation.
Since frequency and wavelength are inversely related by the speed of light (\(c = f\lambda\)), the energy of a photon is also inversely proportional to its wavelength.
Shorter wavelengths (like ultraviolet) correspond to higher energy photons, while longer wavelengths (like infrared) correspond to lower energy photons.
To find the energy in Joules, we must ensure all physical constants and parameters are in standard SI units.
Step 2: Key Formula or Approach:
The energy (\(E\)) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
\(h\) = Planck's constant (\(6.63 \times 10^{-34} \text{ J s}\))
\(c\) = Speed of light (\(3 \times 10^{8} \text{ m/s}\))
\(\lambda\) = Wavelength of the light in meters
Step 3: Detailed Explanation:
Let's prepare our values for the calculation:
Given wavelength, \(\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4 \times 10^{-7} \text{ m}\).
Now, substitute the constants into the equation:
\[ E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^{8} \text{ m/s})}{4 \times 10^{-7} \text{ m}} \]
First, multiply the values in the numerator:
\[ 6.63 \times 3 = 19.89 \]
\[ 10^{-34} \times 10^{8} = 10^{-26} \]
So the numerator is \(19.89 \times 10^{-26}\).
Now, divide by the denominator:
\[ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} \]
\[ E = \left( \frac{19.89}{4} \right) \times 10^{-26 - (-7)} \]
\[ E = 4.9725 \times 10^{-19} \text{ J} \]
Rounding to three significant figures gives us approximately \(4.97 \times 10^{-19} \text{ J}\).
This value is a very tiny amount of energy, which is why we usually don't notice the "graininess" of light in everyday life—there are simply trillions of photons in even a weak light beam.
This energy level corresponds to the violet end of the visible spectrum.
Step 4: Final Answer:
The calculated energy of one photon of 400 nm light is \(4.97 \times 10^{-19} \text{ J}\).
This matches option (A).