Question

what would be the wavelength of photons having the same energy as the electrons in part a ? express your answer in microns.

Answer 1

To find the wavelength of photons having the same energy as the electrons, use the equation λ = hc/E, where E is the energy of the electrons. Plug in the energy from part a and convert the wavelength from meters to microns by dividing by 10^-6. The wavelength of the photons is 1.99 × 10^2 microns.

Step-by-step explanation:

To find the wavelength of photons having the same energy as the electrons, we need to use the equation:

E = hc/λ

Where:

E is the energy of the electrons

h is Planck's constant (6.63 × 10-34 J s)

c is the speed of light (3.00 × 108 m/s)

λ is the wavelength of the photons

Solving for λ, we can rewrite the equation as:

λ = hc/E

Plug in the energy of the electrons from part a to find the wavelength of the photons:

λ = (6.63 × 10-34 J s)(3.00 × 108 m/s)/E

Convert the wavelength from meters to microns by dividing by 10-6:

λ = (6.63 × 10-34 J s)(3.00 × 108 m/s)/E × 10-6 = 1.99 × 102 µm

Therefore, the wavelength of the photons with the same energy as the electrons is 1.99 × 102 microns.

Answer 2

a. the accelerating potential needed to produce electrons with a wavelength of 6.40 nm is approximately
`3.67 * 10^-^2`Volts.

b. the energy of photons having the same wavelength as these electrons is approximately 194 eV.

Part A:

K_e = e * V

K_e is the kinetic energy of the electron

e is the electron charge

V is the accelerating potential

We can also relate the kinetic energy of the electron to its wavelength using the de Broglie wavelength equation:

λ = h / (p * √(2 * K_e / m_e))

λ is the wavelength of the electron

h is Planck's constant

p is the momentum of the electron

m_e is the electron mass

we then solve for V in the first equation and substituting the second equation for K_e:

V = (h² / (2 * e * λ² * m_e))
`^0^.^5`

v =
`3.67 * 10^-^2 Volts.`

Part B:

The energy of photons having the same wavelength as these electrons can be calculated using the equation:

E_photon = h * c / λ

E_photon (eV) = E_photon (J) / e

λ =
`6.40 * 10^-^9` m

h =
`6.62607015 * 10^-^3^4 J s\\c = 2.99792458 * 10^8 m/s\\e = 1.602176634 * 10^-^1^9 C`

E_photon (eV) = 194 eV

complete question:

Part A What accelerating potential is needed to produce electrons of wavelength 6.40 nm? Express your answer in volts. V = 3.67*10-2 V Part B What would be the energy of photons having the same wavelength as these electrons? Express your answer in electron volts. E = 194 eV