9. 2 points SerPSE9 40.P040. My Notes (a) An electron has kinetic energy 2.00 eV

Question

9. 2 points SerPSE9 40.P040. My Notes (a) An electron has kinetic energy 2.00 eV. Find its wavelength. (b) A photon has energy 2.00 eV. Find its wavelength. Need Help? Read It 10. -1 points SerPSE9 40.P.053.WI. My Notes A modified oscilloscope is used to perform an electron interference experiment. Electrons are incident on a pair of narrow slits 0.067 0 ?m apart. The bright bands in the interference pattern are separated by 0.400 mm on a screen 19.0 cm from the slits. Determine the potential difference through which the electrons were accelerated to give this pattern Need Help? Read It tWatch It

Explanation / Answer

(9) First of all convert eV into Joules.

2.0 eV = 2.0 x 1.602 x 10^-19 = 3.204 x 10^-19 J

(a) Kinetic energy of electron = 2.0 eV =  3.204 x 10^-19 J

=> (1/2)*m*v^2 = 3.204 x 10^-19 J

=> v = sqrt[ (2 x 3.204 x 10^-19) / (9.11 x 10^-31)] = 0.839 x 10^6 = 8.39 x 10^5 m/s

Now,

Wavelength = h/mv = (6.62X10^-30Js)/(9.11 x 10^-31kg)(8.39 x 10^5 m/s) = 8.661 x 10^-6 m

Therefore, wavelength of the electron = 8.661 x 10^-6 m = 8.661 microns

(b) Now, take the case of the photon.

E(photon) = 2.0 eV = 3.204 x 10^-19 J

Use the expression -  

E = h f = h c / (wavelength)

=> Wavelength = hc/E

Put the values -

Wavelength = (6.62X10^-30Js)(3 x 10^8m/s)/( 3.204 x 10^-19 J)

= 6.198 x 10^-3 m = 6198 x 10^-6 m = 6198 microns

Therefore, wavelength of the photon = 6198 microns.

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