If an electron having kinetic energ...
Question

# If an electron having kinetic energy ( 2 mathrm{eV} ) is accelerated through the potential difference of 2 Volt. Then calculate the wavelength associated with the electron.

JEE/Engineering Exams
Chemistry
Solution
62
3.7 (3 ratings)

Given conditions ( Rightarrow )
Kinetic Energy of the Electron ( (mathrm{K})=2 mathrm{eV} ). Electron is accelerated through the potential difference of ( 2 mathrm{V} ).
Now, Additional Kinetic Energy ( left(mathrm{K}^{prime}right)=mathrm{eV} ) ( =2 mathrm{eV} )
( therefore ) Total Kinetic Energy ( =2+2 ) ( =4 mathrm{eV} )
( =4 times 1.6 times 10^{-19} mathrm{J} )
( =6.4 times 10^{-19} mathrm{J} )
Now, Using the Formula,
( lambda=frac{h}{sqrt{2 m K}} )
where,
( lambda= ) De-Broglie Wavelength. ( h= ) Plank Constant ( =6.63 times 10^{-34} )
( mathrm{m}= ) Mass of the Electron ( =9.1 times 10^{-31} mathrm{kg} )
( mathrm{K}= ) Kinetic Energy of the Electron ( =6.4 times 10^{-19} mathrm{J} )
( therefore lambda=left(6.63 times 10^{-34}right) div[sqrt{left.left(2 times 9.1 times 10^{-31} times 6.4 times 10^{-19}right)right]} )
( =left(6.63 times 10^{-34}right) div[sqrt{left.left(116.48 times 10^{-50}right)right]} )
( =left(6.63 times 10^{-34}right) divleft[10.79 times 10^{-25}right] )
( =0.614 times 10^{-9} )
( =6.14 times 10^{-10} )
( =6.14 hat{A}left[because 1 hat{A}=10^{-10} mathrm{m}right] )
Hence, the de-Broglie Wavelength of the Electron is 6.14 Angstrom.