Calculate the wavelength of a photon having energy of 1.257 X 10-24 joules. (Planck’s constant is 6.626 x 10-34 joule seconds; the speed of light is 2.998 x 108 m/s)

A.

5.10 m

B.

6.327 m

C.

0.324 m

D.

0.158 m

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Calculate the wavelength of a photon having energy of 1.257 X 10-24 joules. (Planck’s constant is 6.626 x 10-34 joule seconds; the speed of light is 2.998 x 108 m/s)

A.

5.10 m

B.

6.327 m

C.

0.324 m

D.

0.158 m

Answer : The wavelength of a photon is, 3.6 m

Explanation :

[tex]E=\frac{h\times c}{\lambda}[/tex]

where,

E = energy of photon = [tex]5.518\times 10^{-26}J[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]2.998\times 10^{8}m/s[/tex]

[tex]\lambda[/tex] = wavelength of a photon = ?

Now put all the given values in the above formula, we get the wavelength of a photon.

[tex]5.518\times 10^{-26}J=\frac{6.626\times 10^{-34}Js\times 2.998\times 10^{8}m/s}{\lambda}[/tex]

[tex]\lambda=3.599m\approx 3.6m[/tex]

Therefore, the wavelength of a photon is, 3.6 m

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

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:

[tex]E = \dfrac{{hc}}{\lambda }[/tex]

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

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

The wavelength of the photon is [tex]600\,{\text{nm}}[/tex] .

The energy associated with the photon of wavelength [tex]600\,{\text{nm}}[/tex] is:

[tex]\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}[/tex]

The wavelength of photon having energy double of this:

[tex]\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}[/tex]

The new wavelength of the photon will be:

[tex]\lambda ' = \dfrac{{hc}}{{E'}}[/tex]

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

[tex]\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}[/tex]

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

Learn More:

1.Which of the following statements about electromagnetic radiation are true

2.To find the number of neutrons in an atom you would

3.What is the frequency of light for which the wavelength is 7.1*10^2

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.

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

[tex]\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[/tex]

30 nm

For this equation you need to use the equation [tex]E=h\frac{c}{λ}[/tex]

E is your Energy, h is Planck's constant, c is the speed of light and λ is wavelength

First re-arrange the equation to get λ by itself since that's what we are calculating

λ = h (c/E) plug in numbers

λ = 6.626 x10^-34Js [(2.998 x 10^8m/s)/1.257 x10^-24J)] evaluate

λ = 6.626 x10^-34 (2.385 x10^32)

λ ≈ 0.158 m

The frequency of the [tex]\lambda_2 = 622 nm = 622 \cdot 10^{-9} m[/tex] wavelength photon is given by

[tex]f_2 = \frac{c}{\lambda_2}= \frac{3 \cdot 10^8 m/s}{622 \cdot 10^{-9} m}=4.82 \cdot 10^{14} Hz[/tex]

where c is the speed of light.

The energy of this photon is

[tex]E_2=hf_2 = (6.6 \cdot 10^{-34}Js)(4.82 \cdot 10^{14}Hz)=3.18 \cdot 10^{-19} J[/tex]

where h is the Planck constant.

The energy of the first photon is twice that of the second photon, so

[tex]E_1 = 2 E_2 = 2 \cdot 3.18 \cdot 10^{-19}J =6.36 \cdot 10^{-19} J[/tex]

And so now by using again the relationship betwen energy and frequency, we can find the frequency of the first photon:

[tex]f_1 = \frac{E_1}{h}= \frac{6.36 \cdot 10^{-19} J}{6.6 \cdot 10^{-34}Js}=9.64 \cdot 10^{14}Hz[/tex]

and its wavelength is

[tex]\lambda_1 = \frac{c}{f_1}= \frac{3 \cdot 10^8 m/s}{9.64 \cdot 10^{14}Hz} =3.11 \cdot 10^{-7}m = 311 nm[/tex]

So, we see that the wavelength of the first photon is exactly half of the wavelength of the second photon (622 nm).

The wavelength of a photon of energy 5.518 × 10-26 joules is 3.6 m. The energy of the photon is: E = h * f, where h is Planck's constant and f is the frequency. The frequency of the photon is: f = c/λ, where c is the speed of light and λ is the wavelength. Therefore: E = h * f = h * c / λ. We have that E = 5.518 × 10^-26 J; h = 6.626 × 10^-34 Js; c = 2.998 × 10^8 m/s. Substitute this in the formula for the energy of the photon: 5.518 × 10^-26 J = 6.626 × 10^-34 Js * 2.998 × 10^8 m/s / λ. 5.518 × 10^-26 J = 1.986 × 10^-25 Jm / λ. λ = 1.986 × 10^-25 Jm / 5.518 × 10^-26 J. λ = 0.36 × 10 m = 3.6 m.

The wavelength of a photon of energy 5.518 × 10-26 joules is 3.6 m. The energy of the photon is: E = h * f, where h is Planck's constant and f is the frequency. The frequency of the photon is: f = c/λ, where c is the speed of light and λ is the wavelength. Therefore: E = h * f = h * c / λ. We have that E = 5.518 × 10^-26 J; h = 6.626 × 10^-34 Js; c = 2.998 × 10^8 m/s. Substitute this in the formula for the energy of the photon: 5.518 × 10^-26 J = 6.626 × 10^-34 Js * 2.998 × 10^8 m/s / λ. 5.518 × 10^-26 J = 1.986 × 10^-25 Jm / λ. λ = 1.986 × 10^-25 Jm / 5.518 × 10^-26 J. λ = 0.36 × 10 m = 3.6 m.

1.096 i think

Explanation:

[tex]E=h\nu\\\\ \nu=\frac{c}{\lambda} \ \ \ \Rightarrow E=\dfrac{hc}\lambda}\\\\ \lambda=\dfrac{hc}{E}=\dfrac{6,626*10^{-34}Js*2,998*10^{8}\dfrac{m}{s}}{1,257*10^{-24}J}=15,803*10^{\frac{-34+8}{-24}}m=\\\\\\=15,803*10^{-2}m=1,5803*10^{-1}m[/tex]