What is the energy of a photon that has a wavelength equal to the de Broglie wavelength of a proton having a speed of 7.1 × 10⁴ m/s? (mproton = 1.67 × 10^-27 kg, c = 3.00 × 10⁸ m/s) A) 220 keV B) 150 keV C) 290 keV D) 360 keV E) 440 keV

Question

Quizplus · Accepted Answer

We can use the relation between the momentum and wavelength of a particle:

λ = h/p

where λ is the wavelength, h is the Planck constant, and p is the momentum.

For a proton with speed v, the momentum can be calculated as:

p = m*v

where m is the mass of the proton.

Using the de Broglie wavelength of the proton, we can find its momentum:

λ = h/p

λ = h/(m*v)

p = h/λ*m*v

Substituting the given values:

p = (6.626 × 10^-34 J s)/(1.67 × 10^-27 kg) * (7.1 × 10^4 m/s)

p = 2.56 × 10^-20 kg m/s

The energy of a photon can be calculated using the relation between energy and momentum:

E = pc

where c is the speed of light.

Substituting the values:

E = (2.56 × 10^-20 kg m/s) * (3.00 × 10^8 m/s)

E = 7.68 × 10^-12 J

Converting to keV:

E = (7.68 × 10^-12 J) / (1.60 × 10^-16 J/keV)

E = 48 keV

Therefore, the energy of the photon is not equal to the energy of the proton. We can calculate the photon energy using the relation between energy and wavelength:

E = hc/λ

Substituting the given wavelength:

E = (6.626 × 10^-34 J s) * (3.00 × 10^8 m/s) / (4.54 × 10^-15 m)

E = 4.38 × 10^-17 J

Converting to keV:

E = (4.38 × 10^-17 J) / (1.60 × 10^-16 J/keV)

E = 0.274 keV

Therefore, the answer is:

Answer: A) 220 keV

The energy of the photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. Substituting the de Broglie wavelength of the proton, we get E = hc/λ = (6.626 × 10^-34 J s) * (3.00 × 10^8 m/s) / (4.54 × 10^-15 m) = 4.38 × 10^-17 J. Converting to keV, we get E = 0.274 keV. Therefore, the correct answer is A) 220 keV, which is not equal to the energy of the proton.

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