Quantum Mechanics problem.....

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nikhilsigma

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following is a question from Neaman's "Semiconductor Physics and Devices"..

An electron has a kinetic energy of 20meV. Determine the de Broglie wavelength...??
Click to expand...

please explain how to solve this.....
 

λ = h / p

= (6.6 * 10 exp -34) / (5.91 * 10 exp -24) = 1.116 * 10 exp -10 m

= 0.112nm

therefore de borglie wave length of the electron is 0.112nm
 
First of all....how you found the momentum "p" as kinetic energy is given ??

as i know kinetic energy is ½(mv²), and p= mv
BUT v is not given....!!

so this is a problem....... :sad:



Also in book the Answer is : p=7.64x10exp(-26) kgm/s , λ= 86.7 x 10exp(-10) m
 

h is plancks constant h= 6.6* 10exp -34 Joules sec

**broken link removed**
 
try this... the momentum can be determined (non relativistically) as p = sqrt(2*KE*mo) where KE is kinetic energy in Joules (or eV) and mo is the rest mass in kg (or eV/c^2) If using eV/c^2 for your units of mass, it is easier to use the relation p * c = sqrt(2*KE*mo*c^2) Once you know the momentum, getting to the DeBroglie wavelength should be straighforward ... lambda = h / p where h is Planks constant (6.63e-34 Joule*sec)

have Fun
 
good one dude.....I also managed to get that.....

Anyways thanks for the Solution..... ;-)
 

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