Question:
A photon has energy of \(3.1\times10^{-19}\ \text{J}\). Its wavelength (in \(\AA\)) is _ _ _. (round off to one decimal place)
A photon has energy of \(3.1\times10^{-19}\ \text{J}\). Its wavelength (in \(\AA\)) is _ _ _. (round off to one decimal place)
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For photons:
\[
E=\frac{hc}{\lambda}
\]
Higher photon energy corresponds to shorter wavelength.
Updated On: Jun 5, 2026
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Correct Answer: 6387.1
Solution and Explanation
Step 1: Recall the photon energy relation.
Energy of a photon is given by
\[ E=\frac{hc}{\lambda} \] where
\[ E=\text{energy}, \quad h=\text{Planck's constant}, \quad c=\text{speed of light}. \]
Step 2: Rearrange the formula for wavelength.
\[ \lambda=\frac{hc}{E} \]
Step 3: Convert speed of light into SI units.
Given
\[ c=3\times10^{10}\ \text{cm/s} \] Since
\[ 1\ \text{cm}=10^{-2}\ \text{m}, \] \[ c=3\times10^8\ \text{m/s} \]
Step 4: Substitute the values.
\[ \lambda = \frac{(6.6\times10^{-34})(3\times10^8)} {3.1\times10^{-19}} \]
Step 5: Simplify the numerator.
\[ 6.6\times3=19.8 \] Thus,
\[ \lambda = \frac{19.8\times10^{-26}} {3.1\times10^{-19}} \]
Step 6: Calculate the wavelength.
\[ \lambda = 6.387\times10^{-7}\ \text{m} \] Now convert meters into Angstrom units.
\[ 1\ \AA=10^{-10}\ \text{m} \] Therefore,
\[ \lambda = 6.387\times10^{-7}\times10^{10}\ \AA \] \[ =6387.1\ \AA \]
Step 7: Final conclusion.
Hence, the wavelength of the photon is
\[ \boxed{6387.1\ \AA} \]
Energy of a photon is given by
\[ E=\frac{hc}{\lambda} \] where
\[ E=\text{energy}, \quad h=\text{Planck's constant}, \quad c=\text{speed of light}. \]
Step 2: Rearrange the formula for wavelength.
\[ \lambda=\frac{hc}{E} \]
Step 3: Convert speed of light into SI units.
Given
\[ c=3\times10^{10}\ \text{cm/s} \] Since
\[ 1\ \text{cm}=10^{-2}\ \text{m}, \] \[ c=3\times10^8\ \text{m/s} \]
Step 4: Substitute the values.
\[ \lambda = \frac{(6.6\times10^{-34})(3\times10^8)} {3.1\times10^{-19}} \]
Step 5: Simplify the numerator.
\[ 6.6\times3=19.8 \] Thus,
\[ \lambda = \frac{19.8\times10^{-26}} {3.1\times10^{-19}} \]
Step 6: Calculate the wavelength.
\[ \lambda = 6.387\times10^{-7}\ \text{m} \] Now convert meters into Angstrom units.
\[ 1\ \AA=10^{-10}\ \text{m} \] Therefore,
\[ \lambda = 6.387\times10^{-7}\times10^{10}\ \AA \] \[ =6387.1\ \AA \]
Step 7: Final conclusion.
Hence, the wavelength of the photon is
\[ \boxed{6387.1\ \AA} \]
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