answersLogoWhite

0

To find this answer you will have to go through a series of formulas. The first formula you will need to use is the kinetic energy formula (K.E.=1/2mv^2). The mass of an electron is found to be 9.11 x 10^-31. You then divide the mass by two (or multiply by 0.5) and get 4.555 x 10^-31, you will then have to multiply it by your velocity squared, and get your energy in joules. With that energy, you divide by planks constant (6.6 x 10^-34) which eaves you with your frequency. With that very frequency you get the speed of light in air (3 x 10^8) and divide by your frequency which will give you the wavelength needed in meters

User Avatar

Wiki User

16y ago

What else can I help you with?

Related Questions

If proton and an electron have the same speed which has the longer de Broglie wavelength?

It is electron since wavelength = h/(mv), and since proton's mass > electron's mass, electron's wavelength is longer.


What is de broglie's wavelength of electron in meters travelling at half a speed of light?

4.2*10-11


What A typical wavelength for X-radiation is 0.1nm so Calculate the speed of an electron which would have the same wavelength?

To find the speed of an electron with a wavelength of 0.1nm, you can use the de Broglie wavelength formula: λ = h / mv, where λ = wavelength, h = Planck's constant, m = mass of electron, and v = speed of electron. Rearranging the formula to solve for v, we get v = h / (mλ). Plugging in the values (h = 6.63 x 10^-34 J·s, m = 9.11 x 10^-31 kg, and λ = 0.1 x 10^-9 m), you can calculate the speed.


An electron starting from rest accelerates through a potential difference of 388 V What is the final de Broglie wavelength of the electron assuming that its final speed is much less than the spee?

To find the final de Broglie wavelength, you can use the equation λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the electron. The momentum can be calculated as p = √(2mE), where m is the mass of the electron and E is the kinetic energy acquired from the potential difference. Find the final speed of the electron using the equation v = √(2eV/m), where e is the elementary charge. Finally, use the speed to calculate the final momentum and plug it into the de Broglie wavelength formula.


What is the De Broglie wavelength of an electron that strikes the back of the face of a TV screen at 19 the speed of light?

Assuming you mean that the velocity is 1/9th the speed of light then you need to use the de Broglie equation for the wavelength of a particle, which says that the wavelength is equal to Planck's constant divided by the momentum. Thus, λ = h / p = h / (m*v) = h/(m*1/9*c) = 9*h/(m*c) where λ=wavelength, h=Planck's constant, p=momentum, m=mass of the electron, v=velocity, and c=speed of light this gives λ = 9 * 6.626*10^-34 / (9.109*10^-31 * 3.00*10^8) = 2.18*10^-11 meters


What is the characteristic wavelength of the electron when an electron is accelerated through a particular potential field if it attains a speed of 9.38x10 to the power of 6 ms?

The characteristic wavelength of an electron accelerated through a potential field can be calculated using the de Broglie wavelength formula: λ = h / p, where h is the Planck constant and p is the momentum of the electron. Given the speed of the electron, momentum can be calculated as p = m*v, where m is the mass of the electron. Once the momentum is determined, the wavelength can be calculated.


Determine wavelength for electron having velocity 15.0 the speed of light?

Speed of electron as compared to speed of light is: n = 15% c = 299792458 [m/s] v = c*n/100 = 4.5 *10^7 [m/s] So corresponding wavelength as given by the de Broglie equation: h - Planck's constant, m0 - the mass of the electron at zero velocity; lambda = h/p = h/(v*m0) = 6.62606876*10^-34/(4.5 *10^7*9.10938188*10^-31) = 1.61642*10^-11 [m] = 0.16 [angstroms]


How is an electron's wavelength related to its speed and mass?

The wavelength of an electron is inversely proportional to its speed and directly proportional to its mass. This means that as the speed of an electron increases, its wavelength decreases, and as the mass of an electron increases, its wavelength also increases.


Does the de broglie wavelength of a photon become longer or shorter as its velocity increases?

The de Broglie wavelength of a photon remains constant as its velocity increases because a photon always travels at the speed of light in a vacuum. The wavelength of light is determined by its frequency according to the equation λ = c / f.


What is de-Broglie wavelength of an atom at absolute temperature T K?

The de Broglie wavelength of an atom at absolute temperature T K can be calculated using the formula λ = h / (mv), where h is Planck's constant, m is the mass of the atom, and v is the velocity of the atom. At higher temperatures, the velocity of atoms increases, leading to a shorter de Broglie wavelength.


What is de broglie wavelength of one photon?

No, they don't have de-broglie wavelength since this concept is valid for substances having some mass. This concept is mainly used to calculate the wavelength of electrons. The photons are just energy packets and they are not matter and don't have weight. Therefore, they don't have de-broglie wavelength


What is the speed of the electron with a de Broglie wavelength of 235nm?

The DeBroglie wavelength of an electron with 1 eV KE and rest mass energy 0.511 MeV is 1.23 nm. This is around a thousand times smaller than a 1 eV photon. To find the DeBroglie wavelength of an electron, simply divide Planck's constant by the momentum of the electron.