Question

When 400 volt electrons are diffracted by a crystal the angular diffraction pattern is identical with that produced by X–rays of wavelength 0.61Å. Calculate the Planck’s constant.

Answer 1

Correct answer is 6.59 x 10^-34J–s

Answer 2

The relationship between the wavelength of a particle (such as an electron) and its momentum is given by the de Broglie equation:

λ = h / p

where λ is the wavelength, h is Planck's constant, and p is the momentum.

We can use this equation to relate the momentum of the 400 volt electrons to their wavelength:

p = √(2mE)

where m is the mass of an electron, and E is the energy of the electron (given as 400 volts). Substituting this expression for p into the de Broglie equation, we have:

λ = h / √(2mE)

We are told that the angular diffraction pattern of the electrons is identical to that produced by X-rays with a wavelength of 0.61 Å. This implies that the electrons are diffracting at the same angles as the X-rays, and therefore have the same de Broglie wavelength:

λ = 0.61 Å = 6.1 x 10^-11 m

Substituting this into the previous equation and solving for h, we have:

h = λ √(2mE)

h = (6.1 x 10^-11 m) √(2 x 9.109 x 10^-31 kg x 400 eV x 1.602 x 10^-19 J/eV)

h ≈ 6.63 x 10^-34 J-s

Therefore, the value of Planck's constant is approximately 6.63 x 10^-34 J-s.