Question

Obtain the wavelengths of a photon and an electron that have the same energy of 3.0eV.

Answer 1

Given,

Energy of photon, E_photon = 3 eV = 3 * 1.6 * 10^-19 J

Now,

Let wavelength associated with photon be ₁

Energy, E = hc /     Where    h is Plank's constant

                                               c is speed of light

hence

E_photon = hc / ₁

=> 3 * 1.6 * 10^-19 = ((6.624 * 10^-34) * (3 * 10⁸)) / ₁

=> ₁ = ((6.624 * 10^-34) * (3 * 10⁸)) / ( 3 * 1.6 * 10^-19 )

             = 4.14 * 10^-7 = 414 * 10^-9 = 414 nm

Hence, wavelength for photon is 414 nm

Now, for electron

Given

Energy of electron, E_electron = 3 eV = 3 * 1.6 * 10^-19 J

Now,

Let momentum of electron be ₂

Let momentum be p

     mass of electron be m_e

Kinetic energy, K = p² / 2m

hence,

E_electron = p² / 2m_e

=> 3 * 1.6 * 10^-19 = p² / (2 * 9.11 * 10^-31 )

=> p² = ( 3 * 1.6 * 10^-19 ) * (2 * 9.11 * 10^-31 )

=> p² = 87.456 * 10^-50

=> p =    = 9.352 * 10^-25

Now

De Broglie's equation for wavelength of electron is

= h / p           where h is Plank's constant

                                    p is momentum of electron

Now,

=> ₂ = h / p

           = 6.624 * 10^-34 / 9.352 * 10^-25

           = 0.7082 * 10^-9 = 0.71 nm

or ₂ = 0.71 nm

Hence, Wavelength for electron is 0.71 nm