The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.

Wavelength of light of a sodium line, λ = 589 nm = 589 x 10^–9 m

Mass of an electron, m_e= 9.1 x 10^–31 kg

Mass of a neutron, m_n= 1.66 x 10^–27 kg

Planck’s constant, h = 6.6 x 10^-34 Js

Ans (a).

For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:

We have the relation for de Broglie wavelength as:

K = 1/2 m_ev²—————-Eq-1

We have the relation for de Broglie wavelength as :

λ = h /m_ev

Therefore,  v² =h²/λ²m²_e——————— Eq-2
Substituting equation (2) in equation (1), we get the relation:
 

K = 1/2  m_eh²/2λ²m²_e =h²/2λ²m_e—————-Eq-3

= (6.6 x 10^-34)²/2 (589 x 10^–9)² (9.1 x 10^–31 )

≈   6.9 x 10^-25 J

=(6.9 x 10^-25)/(1.6 x 10⁻¹⁹ ) =4.31 x 10 ⁻⁶ eV = 4.31 μ eV

Hence, the kinetic energy of the electron is 6.9 x 10^-25 J or 4.31 μeV.

Ans (b).

Using equation (3), we can write the relation for the kinetic energy of the neutron as:

K= h²/2 λ²m_n

= (6.6 x 10^-34)² /2 (589 x 10^–9)² (1.66 x 10^–27 )

=3.78 x 10⁻²⁸ J

=( 3.78 x 10⁻²⁸ )/(1.6 x 10⁻¹⁹ )

= 2.36 x 10⁻⁹ eV = 2.36 neV

Hence, the kinetic energy of the neutron is 3.78 x 10⁻²⁸ J or 2.36 neV.