Question

The wavelength of the yellow spectral emission line of sodium is 590 nm. At
what kinetic energy would an electron have that wavelength as its de-Broglie
wavelength?

Answer 1

Answer: 4.05 x 10^-29 J

Explanation:

Answer 2

Step by step explanation

Given: Wavelength is $590mn$.

To find: Kinetic energy of an electron.

For calculation of Kinetic energy,

We know that λ $= \frac{h}{p}= \frac{b}{\sqrt{\frac{e}{c^{2} }} - m^{2}_{e} c^{2}}$

So, $K=E - m_{e} c^{2}$

⇒ $K= (\frac{h}{λ}) ^{2} + m^{2} _{e} c^{4} - m_{e} c^{2}$

⇒ $K= (\frac{1240eV nm}{10 x 10^{-3} }) ^{2} + (0.511MeV)^{2} - 0.511MeV$

⇒ $0.015MeV= 15keV$

Using the value $hc=1240eV. nm$

⇒ $E= \frac{1240eV.nm}{10 x 10nm^{-3} }$

⇒ $1.2$ × $10^{5} eV$

⇒ $E=120KeV$

The Kinetic energy of an electron having that wavelength as its de-Broglie wavelength is $120KeV$.