Question

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.

Answer 1

Wavelength of light of a sodium line, λ = 589 nm = 589 × 10⁻⁹ m

Mass of an electron, m_e = 9.1 × 10⁻³¹ kg

Mass of a neutron, m_n = 1.66 × 10⁻²⁷ kg

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

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

`"K" = 1/2 "m"_"e""v"^2` ...............(1)

We have the relation for de Broglie wavelength as:

`lambda = "h"/("m"_"e""v")`

∴ `"v"^2 = "h"^2/(lambda^2"m"_"e"^2)` ........(2)

Substituting equation (2) in equation (1), we get the relation:

`"K" = 1/2 ("m"_"e""h"^2)/(lambda^2"m"_"e"^2) = "h"^2/(2lambda^2"m"_"e")` ..........(3)

= `(6.6 xx 10^(-34))^2/(2 xx (589 xx 10^(-9))^2 xx 9.1 xx 10^(-31))`

≈ 6.9 × 10⁻²⁵ J

= `(6.9 xx 10^(-25))/(1.6 xx 10^(-19))`

= 4.31 × 10⁻⁶ eV

= 4.31 μeV

Hence, the kinetic energy of the electron is 6.9 × 10⁻²⁵ J or 4.31 μeV.

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

`"h"^2/(2lambda^2 "m"_"n")`

= `(6.6 xx 10^(-34))^2/(2 xx (589 xx 10^(-9))^2 xx 1.66 xx 10^(-27))`

= 3.78 × 10⁻²⁸ J

= `(3.78 xx 10^(-28))/(1.6 xx 10^(-19))`

= 2.36 × 10⁻⁹ eV

= 2.36 neV

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