Question

de-broglie wavelength associated with an electron under a potential difference of 100 v​

Answer 1

0.122 nm will be the de-broglie wavelength associated with an electron

Explanation:

when an electron accelerates through the potential difference the kinetic energy is eaual to the eV

= $\frac{mv*v}{2}$ = eV  ..................................................eq 1

where v is velocity of the electron and V is the potential difference

e is the charge on electron= 1.620 * 10 ⁻¹⁹ C

m is the mass of the electron   9.109 * 10⁻³¹ Kg

Using above data in eq 1 we get :

v*v    = 2* (1.620 * 10 ⁻¹⁹ * 100 ) / 9.109 * 10⁻³¹  

   =   35.569*10¹² m/s

    v  = 5.96 * 10⁶ m/s

De broglie wavelength is given by :

λ = $\frac{h}{mv}$ ..................................... eq 2

//  where h is : Planck constant is defined to have the exact value. 6.62607015×10−34 J⋅s.

inserting values in eq 2 we get :

= ( 6.626×10⁻³⁴ ) /  (5.96 * 10⁶ * 9.109 * 10⁻³¹ )

= 0.122 * 10⁻⁹ m

=0.122 nm

Answer 2

Answer:1.2 A° or 1.2 × 10^-10 m

Explanation:

lamda electron = 12.27 ÷√(potential difference)

= 12.27 ÷ √[100]

= 1.2 × 10^-10 m

= 1.2 A°