Question

If an electron having kinetic energy 2 ev is accelerated through the potential difference of 2 volt

Answer 1

Question is little incomplete. You have not clearly mentioned what we have to find. Always in such Question,We need to find the de-Broglie wavelength.

Given conditions ⇒
Kinetic Energy of the Electron (K) = 2 eV.
Electron is accelerated through the potential difference of 2 V.

Now, Additional Kinetic Energy (K') = eV
 = 2 eV
∴ Total Kinetic Energy = 2 + 2
= 4 eV.
= 4 × 1.6 × 10⁻¹⁹ J
= 6.4 × 10⁻¹⁹ J.

Now, Using the Formula,

 λ = $\frac{h}{ \sqrt{2mK} }$
where, 
λ = De-Broglie Wavelength.
h = Plank Constant = 6.63 × 10⁻³⁴
m = Mass of the Electron = 9.1 × 10⁻³¹ kg.
K = Kinetic Energy of the Electron = 6.4 × 10⁻¹⁹ J. 

∴ λ = (6.63 × 10⁻³⁴) ÷ [√(2 × 9.1 × 10⁻³¹ × 6.4 × 10⁻¹⁹)]
 = (6.63 × 10⁻³⁴) ÷ [√(116.48 × 10⁻⁵⁰)]
 = (6.63 × 10⁻³⁴) ÷ [10.79 × 10⁻²⁵]
 = 0.614 × 10⁻⁹
 = 6.14 × 10⁻¹⁰
 = 6.14 Å [∵ 1 Å = 10⁻¹⁰ m]

Hence, the de-Broglie Wavelength of the Electron is 6.14 Angstrom.


Hope it helps.