An electron microscope employs a beam of electrons to obtain an image of an object. What energy must be imparted to each electron of the beam to obtain a wavelength of 15.2 pm? Obtain the energy in electron volts (eV) (1 eV = 1.602 10-19 J).

Question

An electron microscope employs a beam of electrons to obtain an image of an object. What energy must be imparted to each electron of the beam to obtain a wavelength of 15.2 pm? Obtain the energy in electron volts (eV) (1 eV = 1.602 10-19 J).

Answer 1

E = hc/wavelength

Plug in the wavelength, calculate E (in Joules) and convert to eV.

Answer 2

firstly find the speed of the electron by using the formula wavelength =Planck's constant / mass * velocity( speed) as the speed of light does not apply in this case as electrons are slower and not part of EMR. make speed the subject of the formula. after finding speed use the equation E=hv/ wavelength. after obtaining the energy in joules convert the joules to eV as required by the question.

mass of an electron is equal to 9.11 *10^-31 kg
planck's constant =6.626 *10^-34
wavelength given is 10.0 pm =1*10*-11 m
final answer should be 3eV

Answer 3

To determine the energy required for each electron, we can use the de Broglie equation, which relates the wavelength (λ) of a particle to its momentum (p):

λ = h / p

Where:
λ = wavelength
h = Planck's constant (6.626 x 10^-34 J·s)
p = momentum

Since electrons are considered to have relativistic mass, we need to use the relativistic momentum equation:

p = γ * m * v

Where:
p = momentum
γ = Lorentz factor (γ = 1 / √(1 - (v^2 / c^2)))
m = rest mass of the electron (9.11 x 10^-31 kg)
v = velocity of the electron
c = speed of light (3 x 10^8 m/s)

Since the electron's velocity is difficult to measure directly, we can approximate it using the kinetic energy:

K = (1/2) * m * v^2

Where:
K = kinetic energy

To calculate the energy required for each electron, we need to find the velocity using the kinetic energy formula:

v = √((2 * K) / m)

Substituting this into the relativistic momentum equation, we get:

p = γ * m * √((2 * K) / m)

Now, we can substitute this momentum value into the de Broglie equation:

λ = h / (γ * m * √((2 * K) / m))

Since we know the desired wavelength (λ = 15.2 pm = 15.2 x 10^-12 m), we can rearrange the equation:

K = ((h^2) / (γ^2 * m * λ^2)) / 2

Finally, we can substitute values and solve for the energy (K) in joules:

K = ((6.626 x 10^-34 J·s)^2) / (γ^2 * (9.11 x 10^-31 kg) * (15.2 x 10^-12 m)^2) / 2

After calculating K in joules, we can convert it to electron volts (eV) using the conversion factor:

1 eV = 1.602 x 10^-19 J

By multiplying the value of K in joules by (1 eV / (1.602 x 10^-19 J)), we can obtain the energy in electron volts (eV).

Answer 4

To find the speed of the electron, we can rearrange the formula:

wavelength = Planck's constant / (mass * velocity)

Rearranging for velocity:

velocity = Planck's constant / (mass * wavelength)

Plugging in the values:

velocity = (6.626 * 10^-34) / (9.11 * 10^-31 * 15.2 * 10^-12)
= 45655.8 m/s

Now we can use the equation E = (Planck's constant * velocity) / wavelength to calculate the energy:

E = (6.626 * 10^-34 * 45655.8) / (15.2 * 10^-12)
= 1.988 * 10^-25 J

To convert this to electron volts (eV), we can use the conversion factor:

1 eV = 1.602 * 10^-19 J

Converting:

Energy in eV = (1.988 * 10^-25) / (1.602 * 10^-19)
= 1.24 * 10^-6 eV

So the energy imparted to each electron of the beam is approximately 1.24 * 10^-6 eV.