Question:
The energy associated with radiation of wavelength 700 nm is (in eV):
The energy associated with radiation of wavelength 700 nm is (in eV):
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Energy of photon is inversely proportional to wavelength:
\[
E(\text{eV}) \approx \frac{1240}{\lambda(\text{nm})}
\]
Updated On: Jul 18, 2026
- 1.77
- 17.7
- 4.3
- 14.3
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The Correct Option is A
Solution and Explanation
Step 1: Understand the basic formula for photon energy.
The energy of a photon is given by Planck’s equation: \[ E = \frac{hc}{\lambda} \] where \(h\) is Planck’s constant, \(c\) is speed of light, and \(\lambda\) is wavelength. This formula gives energy in joules, which we later convert into electron volts (eV).
Step 2: Substitute standard values.
We take: \[ h = 6.626 \times 10^{-34} \, \text{J s}, \quad c = 3 \times 10^8 \, \text{m/s} \] Given wavelength: \[ \lambda = 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m} \] Substituting into formula: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} \]
Step 3: Simplify the numerical expression.
First multiply numerator: \[ 6.626 \times 3 = 19.878 \times 10^{-34+8} = 19.878 \times 10^{-26} \] Now divide by \(700 \times 10^{-9}\): \[ E = \frac{19.878 \times 10^{-26}}{7 \times 10^{-7}} \] This gives: \[ E \approx 2.8397 \times 10^{-19} \, \text{J} \]
Step 4: Convert joules into electron volts.
We use: \[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \] So, \[ E = \frac{2.8397 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.77 \, \text{eV} \]
Step 5: Interpret the physical meaning.
This energy corresponds to visible light radiation in the red region of the spectrum. Longer wavelength implies lower energy, consistent with inverse proportionality between energy and wavelength.
Step 6: Final conclusion.
Thus, the energy of radiation of wavelength 700 nm is: \[ \boxed{1.77 \, \text{eV}} \]
The energy of a photon is given by Planck’s equation: \[ E = \frac{hc}{\lambda} \] where \(h\) is Planck’s constant, \(c\) is speed of light, and \(\lambda\) is wavelength. This formula gives energy in joules, which we later convert into electron volts (eV).
Step 2: Substitute standard values.
We take: \[ h = 6.626 \times 10^{-34} \, \text{J s}, \quad c = 3 \times 10^8 \, \text{m/s} \] Given wavelength: \[ \lambda = 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m} \] Substituting into formula: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} \]
Step 3: Simplify the numerical expression.
First multiply numerator: \[ 6.626 \times 3 = 19.878 \times 10^{-34+8} = 19.878 \times 10^{-26} \] Now divide by \(700 \times 10^{-9}\): \[ E = \frac{19.878 \times 10^{-26}}{7 \times 10^{-7}} \] This gives: \[ E \approx 2.8397 \times 10^{-19} \, \text{J} \]
Step 4: Convert joules into electron volts.
We use: \[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \] So, \[ E = \frac{2.8397 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.77 \, \text{eV} \]
Step 5: Interpret the physical meaning.
This energy corresponds to visible light radiation in the red region of the spectrum. Longer wavelength implies lower energy, consistent with inverse proportionality between energy and wavelength.
Step 6: Final conclusion.
Thus, the energy of radiation of wavelength 700 nm is: \[ \boxed{1.77 \, \text{eV}} \]
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