Question

What is the wavelength of a photon whose energy is twice that of a photon with a 600 nm wavelength?

Answer 1

Planck's equation states that
E = hf
where
E =  the energy,
h = Planck's constant
f =  the frequency

Because
c = fλ
where
c =  velocity of light,
λ = wavelength
therefore
E = h(c/λ)

Photon #1:
The wavelength is λ₁ = 60 nm.
The energy is
E₁ = (hc)/λ₁

Photon #2:
The energy is twice that of photon #1, therefore its energy is
E₂ = 2E₁ = (hc)/λ₂.

Therefore
$\frac{E_{2}}{E_{1}}= \frac{(hc)/\lambda_{2}}{(hc)/60 \, nm} =2\\ \frac{60}{\lambda_{2}} =2 \\ \lambda_{2} = \frac{60}{2} =30 \, nm$

Answer:  30 nm

Answer 2

The wavelength of the photon having twice the energy as that of the photon of wavelength $600\,{\text{nm}}$ is $\boxed{300\,{\text{nm}}}$ .

Further Explanation:

The photons are the small packets of energy that move at the speed of light. The photons are considered to remain always in motion. The energy associated with a moving photon is given by:

$E = \dfrac{{hc}}{\lambda }$

Here,  $E$  is the energy associated with the photon, $h$ is the Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of the moving photon.

The value of the Planck’s constant is $6.6 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}$ .

The wavelength of the photon is $600\,{\text{nm}}$ .

The energy associated with the photon of wavelength $600\,{\text{nm}}$ is:

$\begin{aligned}{E_1}&=\frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{600 \times {{10}^{ - 9}}}}\\&=\frac{{1.98 \times {{10}^{ - 25}}}}{{6 \times {{10}^{ - 7}}}}\\&= 3.3 \times {10^{ - 19}}\,{\text{J}}\\\end{aligned}$

The wavelength of photon having energy double of this:

$\begin{aligned}E' &= 2{E_1}\\&= 2 \times\left( {3.3 \times {{10}^{ - 19}}} \right)\,{\text{J}}\\&{\text{ = 6}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 19}}\,{\text{J}}\\\end{aligned}$

The new wavelength of the photon will be:

 $\lambda ' = \dfrac{{hc}}{{E'}}$

Substitute $6.6 \times {10^{ - 19}}\,{\text{J}}$ for $E'$ in above expression.

$\begin{aligned}\lambda ' &= \frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{6.6 \times {{10}^{ - 19}}}}\\&=\frac{{1.98 \times {{10}^{ - 25}}}}{{6.6 \times {{10}^{ - 19}}}}\,{\text{m}}\\&= 3.0 \times {10^{ - 7}}\,{\text{m}}\\&= 300\,{\text{nm}}\\\end{aligned}$

The wavelength of the photon having twice the energy as that of the photon of wavelength $600\,{\text{nm}}$ is $\boxed{300\,{\text{nm}}}$.

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Photon and Energy

Keywords:  Wavelength, photon, energy, E=hc/lamda, 600nm, twice the energy, Planck’s constant, small packets of energy, 300nm, speed of light.