Correct Answer - Option 3 : λ =
\(\frac{0.286}{\sqrt{V}}\)Å
CONCEPT:
Dual nature of matter:
- According to de Broglie, the matter has a dual nature of wave-particle.
-
The wave associated with each moving particle is called matter waves.
- de Broglie wavelength associated with the particle
\(⇒ λ = \frac{h}{mv}\)
Where, h = Planck's constant, m = mass of a particle and v = velocity of a particle
Work-Energy Theorem:
-
The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.
⇒ W = ΔKE
Where W = work done and ΔKE = change in kinetic energy
CALCULATION:
Given V volts = potential difference, charge of a proton e = 1.6×10-19 C and m = 1.67×10-31 kg
- We know that if a charge of e coulombs is moved through a potential difference of V volts then the work done is given as,
⇒ W = eV -----(1)
- By work-energy theorem, the work done in accelerating a proton from rest through the electric field will be equal to the change in kinetic energy of the proton,
⇒ W = ΔKE
⇒ W = KE2 - KE1 -----(2)
Where KE1 = initial kinetic energy and KE2 = final kinetic energy
Initially, the proton is at rest so,
⇒ KE1 = 0
Final kinetic energy is given as,
\(⇒ KE_2=\frac{1}{2}× m v^2\)
So,
\(⇒ W=\frac{1}{2}× m v^2-0\)
\(⇒ W=\frac{1}{2}× m v^2\) -----(3)
By equation 1 and equation 3,
\(⇒ eV=\frac{1}{2}× m v^2\)
\(⇒ v=\sqrt{\frac{2eV}{m}}\) -----(4)
- The de Broglie wavelength associated with the proton is given as,
\(⇒ λ = \frac{h}{mv}\) -----(5)
By equation 4 and equation 5,
\(⇒ λ = \frac{h}{m}×\sqrt{\frac{m}{2eV}}\)
\(⇒ λ = \frac{h}{\sqrt{2meV}}\)
Since h = 6.6 × 10-34 J-sec
\(⇒ λ = \frac{6.6×10^{-34}}{\sqrt{2×1.67×10^{-27}×1.6×10^{-19}×V}}\)
\(\Rightarrow \lambda=\frac{0.286\times10^{-10}}{\sqrt{V}}\,m\)
\(\Rightarrow \lambda=\frac{0.286}{\sqrt{V}}\,Å\)
- Hence, option 3 is correct.
