0
What is the speed of the electron with a de Broglie wavelength of 235nm?
Wiki User
∙ 8y ago
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.
Wiki User
∙ 9y ago
Wiki User
∙ 8y agoIn order to solve this we will utilize the following 2 equations and 2 physical constants to perform such a calculation,
[1] Electron's momentum (P) = Mass of electron (m) * Velocity of electron (V)
P = m*V
[2] De Broglie wavelength (λ) = [ Planck's constant (h)] / [ Electron's Momentum (P)]
λ = h / P
1)Planck's constant, h = 6.63 * 10^(-34) (kg*m^2)/s
2)Electron rest mass, m = 9.11 * 10^(-31) kg
STEP 1) Using equation [2], solve for the electron's momentum (P) using the given wavelength, 235 nano meters
P = h / λ = [ 6.63 * 10^(-34) (kg*m^2)/s ] / [ 235 * 10^(-9) m ] = 2.8213 * 10^(-27) (kg*m)/s
STEP 2) Using equation [1], solve for the velocity of the electron (V) using the answer for momentum (P) from STEP 1
V = P / m = [ 2.8213 * 10^(-27) (kg*m)/s ] / [ 9.11 * 10^(-31) kg ] = 3097 m/s (approx.)
Therefore the velocity of an electron with a de Broglie wavelength of 235nm is:
approximately 3097 m/s
- NOTE**
Wiki User
∙ 9y agoBroglie Equation : Lamda=h/p
h is Plank's constant= 6.626*10^34 J/s
P (momentum)= mass *velocity
mass must be in Kg and Velocity in meters/second
Lamda represent wavelength is meters per second.
10^-9 = nano
Wiki User
∙ 10y agoVelocity = 5.2 X 105 m/s
Mass = 9.11 X 10-31 kg
Momentum p = 4.7372 X 10-25 kg m/s
De Broglie Wavelength = h/p = 1.399 X 10-9 m = 1.4 nm
Add your answer:
Earn +20 pts
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.
How do you calculate the speed of an electron using the de Broglie wavelength?
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
Because c the speed of electromagnetic radiation is a constant the wavelength of the radiation is....?
The wavelength is inversely proportional to its frequency. That is, as the frequency increases, the wavelength decreases and vice versa.
Velocity of a wave increases and the wavelength stays the same. What will the result be when this happens?
Assuming an electromechanical wave not much. The speed of the wave depends on the medium that the wave is passing through. In a vacuum it is the speed of light, through something else a lesser speed. The wavelength stays the same and the frequency stays the same.
What is the wavelength of a photon with an energy of 3.26 10 19 J?
The wavelength is 610 nm.
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 is de-Broglie wavelength of an atom at absolute temperature T K?
de Broglie wavelength depends only on the mass and speed of the particle and not on the temperature
Why is de broglie wavelength associated with macroscopic objects is not observed in daily life?
Because such a wavelength is way too small to be significant. The de Broglie wavelength is inversely proportional to an object's momentum (mass x 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?
An electron, starting from rest, accelerates through a potential difference of 417 V.
Who stated that matter behaves as both waves and particles?
Louis de Broglie
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
How do you calculate the speed of an electron using the de Broglie wavelength?
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
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 to calculate the wavelength?
This question is from Bohr's atomic model. The total length of the orbit is an integral multiple of the wavelength of an electron. The relation given by 2(pi)(radius)=n(wavelength), where n is the principal quantum number. Proof of this came later from De-Broglie's hypothesis, (wavelength)=h/(linear momentum) It is- (wavelength)=h/mv .....I From Bohr's model (Quantization of angular momentum), mvr=nh/2(pi) So, 2(pi)r=n(h/mv) From I, 2(pi)r=n(wavelength)
What does the wavelength of electron if it moves with the speed of light?
-- First of all, since the electron has rest mass, it can never move at the speedof light.-- Following DeBroglie, the electron's wavelength is such that an integral numberof them fit around the length of the electron's orbit when it's bound to an atom.
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?
A characteristic wavelength of an electron can be known as the DeBroglie Wavelength. It is a formula in physics which relays energy and momentum.
