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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.
Solution
Wavelength of light of a sodium line, λ = 589 nm = 589 × 10−9 m
Mass of an electron, me = 9.1 × 10−31 kg
Mass of a neutron, mn = 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.
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